Hartshorne II.1

1.1. Let $\mathscr{F}$ be the sheaf described above. Observe that $\mathscr{F}_P=A$ for each $P\in X.$ Therefore the sheafification of $\mathscr{F}$ is the sheaf such that $\mathscr{F}(U)$ is the set of continuous functions from $U\to \bigcup_{P\in X}\mathscr{F}_P.$ Note $\mathscr{F}_P=A.$ Suppose that $f:U\to\bigcup_{P\in X}\mathscr{F}_P$ is a function such that for each $P\in U$ there exists a neighborhood $V_P$ of $P$ such that $t^P\in \mathscr{F}(V_P)=A$ with $t^P_Q=f(Q)$ for all $Q\in V_P.$ But $t^P_Q=t^P\in A.$ Thus in fact $f$ is a continuous function from $U\to A$ and addition works as wexpected. So $\mathscr{F}$ is the constant sheaf described in example 1.0.3.

1.2.a. Let $s_P\in(\text{ker}(\varphi))_P.$ Then there exists open $U$ containing $P$ and $s\in\mathscr{F}(U)$ with germ $s_P$ such that $\varphi(U)(s)=0.$ Thus $\varphi(s)_P=\varphi_P(s_P)=0.$ So $s_P\in\text{ker}(\varphi_P).$ Next suppose that $s_P\in\text{ker}(\varphi_P).$ Then for some open neighborhood $U$ of $P$ we must have $s\in\mathscr{F}(U)$ with $s|_P$ and hence $\varphi(s)\in\mathscr{G}(U)$ such that $\varphi(s)|_P=\varphi_P(s)=0.$ Thus there exists a neighborhood $V\subseteq U$ of $P$ such that $\varphi(s)|_V=0.$ So $\varphi(s|_V)=\varphi(s)|_V=0$ and $(s|_V)_P=s_P$ and therefore $s_P\in(\text{ker}(\varphi))_P.$

The argument follows similarly for images, or see the sheafification construction which explicitly constructs $\text{im}(\varphi)$ such that $\text{im}(\varphi)_P=\text{im}(\varphi_P).$

1.2.b. Note that $\text{ker}(\varphi)=0$ if and only if $\text{ker}(\varphi)_P=0$ for all $P\in X.$ By part (a), this is true if and only if $\text{ker}(\varphi_P)=0$ for all $P,$ i.e., if the morphism is injective on the stalks.

Let $\varphi:\mathscr{F}\to\mathscr{G}$ be a morphism of sheaves. If $\text{im}(\varphi)=\mathscr{G}$ then $\text{im}(\varphi)_P=\mathscr{G}_P$ for all $P.$ Moreover, $\text{im}(\varphi_P)=\text{im}(\varphi)_P$ by part (a). Suppose conversely that $\text{im}(\varphi_P)=\mathscr{G}_P$ for all $P\in X.$ Suppose that $U\subseteq X$ and $s\in\mathscr{G}(U).$ Then $s_P\in\text{im}(\varphi_P)$ for each $P\in U.$ So there exist $V_p\subseteq U$ and $s^p\in\text{im}(\varphi)(V_p)$ such that $s^p=s|_{V_P}.$ But then by the sheaf axiom, $s\in\text{im}(\varphi)(U).$ So $\text{im}(\varphi)=\mathscr{G}.$

1.2.c. This follows immediately from parts (a) and (b).

Lemma 1. Suppose that $\mathscr{F}\subseteq\mathscr{G}$ with $\mathscr{F}$ a presheaf and $\mathscr{G}$ a sheaf. We define by $\overline{\mathscr{F}}(U)=\left\{s\in\mathscr{G}(U) : \exists\text{ a cover } \{V_i\} \text{ of } U \text{ and } s_i\in\mathscr{F}(V_i), s|_{V_i}=s_i\right\}.$ Then $\overline{\mathscr{F}}=\mathscr{F}^+$ where $\mathscr{F}^+$ is the sheafification of $\mathscr{F}.$

It is easy to see that $\overline{\mathscr{F}}$ is a sheaf. By the universal property of the sheafification of $\mathscr{F},$ denoted $\mathscr{F}^+,$ we have morphisms $f:\mathscr{F}\to\mathscr{F}^+$ and $\psi:\mathscr{F}^+\to\overline{\mathscr{F}}$ such that the inclusion map $i:\mathscr{F}\to\overline{\mathscr{F}}$ factors as $i=\psi\circ f.$ On the other hand, given $s\in\overline{\mathscr{F}}(U)$ we may choose a cover $\{V_i\}$ of $U$ and $s_i\in\mathscr{F}(V_i)$ with $s|_{V_i}=s_i.$ There also exists $t\in F^+$ such that $t|_{V_i}=s_i.$ Thus we may define $\beta:\overline{\mathscr{F}}(U)\to\mathscr{F}^+(U)$ by $s\mapsto t.$ This gives us a morphism $\beta:\overline{\mathscr{F}}\to\mathscr{F}^+$ and $\beta$ and $\psi$ are inverses.

Corollary 2. Suppose that $f:\mathscr{F}\to\mathscr{G}$ is a morphism of sheaves. Then $\text{im}(f)\subseteq\mathscr{G}.$

1.3.a. By the lemma above we see that the sheafification of the image presheaf of $\varphi,$ regarded as a subsheaf of $\mathscr{G},$ is equal to $\mathscr{G}$ if and only if the condition is satisfied.

1.3.b. Let $X=\{P,Q,R\}$ with nontrivial open sets $U_P=\{P,R\},$ $U_Q=\{Q,R\},$ and $U_R=\{R\}.$ Take $\mathscr{F}$ to be the constant sheaf of $\mathbb{Z}$ on $X$ and $\mathscr{G}$ to be $\mathbb{Z}_P\oplus\mathbb{Z}_Q$ where $\mathbb{Z}_P,\mathbb{Z}_Q$ are the skyscraper sheaf over $P$ and $Q$ respectively. Let us define $\varphi:\mathscr{F}\to\mathscr{G}$ such that $\varphi(X):\mathscr{F}(X)\to\mathscr{G}(X)$ is the map $z\mapsto (z,z),$ and the rest of the maps are induced by restrictions. Observe that this morphism is surjective since given $s=(z_1,z_2)\in\mathscr{G}(X)$ we have $z_1\in\mathscr{F}(U_P)$ and $z_2\in\mathscr{F}(U_Q)$ such that $\varphi(U_P)(z_1)=s|_{U_P}$ and $\varphi(U_Q)(z_2)=s|_{U_Q}.$ Thus $\varphi$ is a surjective morphism despite the fact that $\varphi(X)$ is not surjective.

Accreditation. The idea for this example comes from stackexchange.

1.4.a. By exercise 1.2.a it suffices to show injectivity on the stalks. Observe that the maps $\theta_1:\mathscr{F}\to\mathscr{F}^+$ and $\theta_2:\mathscr{G}\to\mathscr{G}^+$ are bijective on stalks by definition. The map $\varphi:\mathscr{F}\to\mathscr{G}$ is also injective on stalks. Thus the map $\theta_2\circ\varphi:\mathscr{F}\to\mathscr{G}^+$ is injective on stalks and thus the induced map $\varphi^+:\mathscr{F}^+\to\mathscr{G}^+$ is injective on stalks as $\varphi^+\circ\theta_1=\theta_2\circ\varphi.$

1.4.b. Already showed in Lemma 1 and Corollary 2 earlier.

1.5. If a morphism $\varphi:\mathscr{F}\to\mathscr{G}$ is bijective, then $\varphi(U):\mathscr{F}(U)\to\mathscr{G}(U)$ is bijective for each $U\subseteq X$ (see Eisenbud exercise I-9). Thus $\varphi(U)$ is a group (ring, module, etc.) isomorphism for each $U.$ Thus we may define an inverse morphism $\varphi^{-1}$ by taking $\varphi^{-1}(U)=(\varphi(U))^{-1}.$ It easily follows that this is a morphism of sheaves and it is clearly the inverse of $\varphi.$

1.6.a. Observe that the map $\mathscr{F}\to\mathscr{F}/\mathscr{F}&39;$ given locally by $s\mapsto s+\mathscr{F}&39;(U)$ for $s\in\mathscr{F}(U)$ is surjective onto the quotient presheaf. Thus it is surjective onto the quotient sheaf as the image sheaf of this map and quotient sheaf $\mathscr{F}/\mathscr{F}&39;$ are both the sheafification of the the quotient presheaf.

1.6.b. Since $\mathscr{F}&39;\hookrightarrow\mathscr{F}$ we may identify $\mathscr{F}&39;$ with a subsheaf of $\mathscr{F}.$ Thus by (a) we have $\mathscr{F}&39;&39;\cong\mathscr{F}/\mathscr{F}&39;.$

Lemma 3. Suppose that $\mathscr{F}$ is a sheaf, $\mathscr{P}$ is a presheaf, and $\varphi:\mathscr{F}\to\mathscr{P}$ is a map of presheaves. Then we naturally have a map of sheaves $\varphi^+:\mathscr{F}\to\mathscr{P}^+$ and $\text{ker}(\varphi^+)=(\text{ker}^p(\varphi))^+$ where $\text{ker}^p$ denotes the presheaf kernel.

The map $\varphi^+:\mathscr{F}\to\mathscr{P}^+$ is simply given by the composition $\theta\circ\varphi$ where $\theta:\mathscr{P}\to\mathscr{P}^+$ is the unique map given by sheafification. It is immediate that $\text{ker}^p(\varphi)\subseteq\text{ker}(\varphi^+)$ and hence $(\text{ker}^p(\varphi))^+\subseteq\text{ker}(\varphi^+).$ Thus it remains to show the reverse inclusion.

Suppose that $s\in\text{ker}(\varphi^+)(U).$ Then $\varphi^+(s)_P=0$ for all $P\in U.$ By the definition of $\mathscr{P}^+$ and $\theta,$ this implies that $\varphi(s)_P=0$ for all $P\in U$ (recall $\theta$ sends a section $t\in\mathscr{P}(U)$ to the function $U\to\bigcup_{P\in X}\mathscr{P}_P$ given by $Q\mapsto t_Q$ ). This implies that for each $P\in U$ there exists an open neighborhood $V_P$ of $P$ in $U$ such that $\varphi(s)|_{V_P}=0.$ This implies that $s|_{V_P}\in\text{ker}^p(\varphi)$ for each $P\in U.$ Therefore $s\in(\text{ker}^p(\varphi))^+$ as it is the result of gluing the collection of sections $\{s|_{V_P}\}$ over the open cover $\{V_P\}.$

1.7.a. Observe that by Exercise 1.6.a we have an exact sequence $0\to\text{ker}(\varphi)\to\mathscr{F}\to\text{im}(\varphi)\to 0.$ Thus by Exercise 1.6.b we have $\text{im}\varphi\cong\mathscr{F}/\text{ker} \varphi.$

1.7.b. Let $\text{im}^{p},\text{coker}^{p}$ be the presheaf image and cokernel, respectively. By definition, for each open $U\subseteq X$ we have an exact sequence $0\to\text{im}^{p}(\varphi)(U)\to\mathscr{G}(U)\to\text{coker}^{p}(\varphi)(U)\to 0,$ and hence an “exact sequence of presheaves” $0\to\text{im}^{p}(\varphi)\to\mathscr{G}\to\text{coker}^{p}(\varphi)\to 0.$ By exercise 1.4.a we see that $0\to\text{im}(\varphi)\to\mathscr{G}$ is exact after sheafifying $\text{im}^p(\varphi).$ Finally, by Lemma 3 above, we have that the sheafification of $\text{im}^p$ is the kernel of the natural map from $\mathscr{G}\to\text{coker}(\varphi).$ Thus we have the exact sequence $0\to\text{im}(\varphi)\to\mathscr{G}\to\text{coker}(\varphi)\to 0.$ Part (a) therefore implies that $\text{coker}(\varphi)\cong\mathscr{G}/\text{im}(\varphi).$

1.1.8. Suppose that $0\to\mathscr{F}&39;\xrightarrow{\varphi}\mathscr{F}\xrightarrow{\psi}\mathscr{F}&39;&39;$ is an exact sequence of sheaves on $X.$ In order to show that $\Gamma(U,\cdot)$ is right exact, it suffices to show that $0\to\mathscr{F}&39;(U)\xrightarrow{\varphi(U)}\mathscr{F}(U) \xrightarrow{\psi(U)}\mathscr{F}&39;&39;(U)$ is exact for each open $U\subseteq X.$ Equivalently, we must show that $\varphi(U)$ is injective and that $\text{im}(\varphi(U))=\text{ker}(\psi(U)).$ Observe that since $\varphi$ is injective, it immediately follows (from Eisenbud-Harris exercise I-9) that $\varphi(U):\mathscr{F}&39;(U)\to\mathscr{F}(U)$ is injective for each $U\subseteq X.$ Thus $0\to\mathscr{F}&39;(U)\to\mathscr{F}(U)$ is exact for each $U.$

It remains to show that $\text{im}(\varphi(U))=\text{ker}(\psi(U))$ for any opens subset $U.$ Observe that by the exact sequence of sheaves we have that $\text{im}(\varphi)=\text{ker}(\psi)$ (by Corollary 2 above we may view $\text{im}(\varphi)$ as a subsheaf of $\mathscr{F}$ and then the surjectivity of the maps $\varphi_x:\mathscr{F}_x&39;\to\text{ker}(\psi)_x$ gives us that $\text{im}(\varphi)=\text{ker}(\psi)$ ). Thus the morphism $\varphi|_{\mathscr{F}&39;}:\mathscr{F}&39;\to\text{ker}(\psi)$ is surjective. Moreover, $\varphi$ is bijective since $\text{ker}(\varphi)=0.$ Therefore $\varphi|_{\mathscr{F}&39;}(U):\mathscr{F}&39;(U)\to\text{ker}(\psi)(U)$ is bijective for each $U$ (Eisenbud exercise I-9).

1.9. Products. Clearly $\mathscr{F}\oplus\mathscr{G}$ is a presheaf with restriction maps given by the direct product of the restriction maps on $\mathscr{F},\mathscr{G}.$

Suppose that $(s_1,s_2)\in\mathscr{F}\oplus\mathscr{G}(U)$ and $(t_1,t_2)\in\mathscr{F}\oplus\mathscr{G}(V)$ such that $(s_1,s_2)|_{U\cap V}=(t_1,t_2)|_{U\cap V}.$ Then $s_i|_{U\cap V}=t_i|_{U\cap V},$ $i=1,2,$ and thus there exists a unique $r_1\in\mathscr{F}(U\cup V)$ and $r_2\in\mathscr{G}(U\cup V)$ such that $r_i|_U=s_i$ and $r_i|_V=t_i.$ Thus $\mathscr{F}\oplus\mathscr{G}$ satisfies the sheaf axiom.

It remains to show that $\mathscr{F}\oplus\mathscr{G}$ satisfies the universal properties of direct sum and direct product. Let $i_1,i_2$ be the inclusion moprhims and $p_1,p_2$ the projection moprhims of $\mathscr{F},\mathscr{G}$ respectively.

Suppose $\mathscr{H}$ is a sheaf and $\varphi_1:\mathscr{H}\to\mathscr{F}$ and $\varphi_2:\mathscr{H}\to\mathscr{G}$ morphisms. Then observe that $\psi:\mathscr{H}\to\mathscr{F}\oplus\mathscr{G}$ defined by $\psi(U)=(i_1\circ\varphi_1)(U)\oplus (i_2\circ\varphi_2)(U)$ is a map satisfying $p_1\circ\psi=\varphi_1$ and $p_2\circ\psi=\varphi_2.$

Suppose $\mathscr{H}$ is a sheaf and $\varphi_1:\mathscr{F}\to\mathscr{H}$ and $\varphi_2:\mathscr{G}\to\mathscr{H}$ are morphisms. Then $\psi:\mathscr{F}\oplus\mathscr{G}\to\mathscr{H}$ defined by $\psi(U)=\varphi_1(U)\oplus\varphi_2(U)$ is a morphism satisfying $\psi\circ i_1=\varphi_1$ and $\psi\circ i_2=\varphi_2.$

1.10. Direct limits. Given the directed system $g_i:\mathscr{F}_i\to\mathscr{G},$ we locally have maps $f_i(U):\mathscr{F}_i(U)\to\varinjlim\mathscr{F}_i(U)$ and a map $\varphi(U):\varinjlim\mathscr{F}_i(U)\to\mathscr{G}(U)$ such that $g_i(U)=\varphi(U)\circ f_i(U).$ These define morphisms of presheaves showing that the presheaf $U\mapsto\varinjlim\mathscr{F}_i(U)$ is a direct limit in the category of presheaves. Thus the sheafification $\varinjlim \mathscr{F}_i$ is a direct limit in the category of sheaves.

1.11. Suppose that $s\in\varinjlim\mathscr{F}_i(U),$ $t\in\varinjlim\mathscr{F}_i(V)$ such that $s|_{U\cap V}=t|_{U\cap V}.$ Then by the definition of direct limit, there exists a $j$ such that $s&39;\in\mathscr{F}_j(U),$ $t&39;\in\mathscr{F}_j(V)$ with $s&39;\mapsto s,$ $t&39;\mapsto t,$ and $s&39;|_{U\cap V}=t&39;|_{U\cap V}.$ Thus by the sheaf axiom on $\mathscr{F}_j$ there esists $r&39;\in\mathscr{F}_j(U\cup V)$ such that $r&39;|_U=s&39;$ and $r&39;|_V=t&39;.$ Therefore $r&39;\mapsto r\in\varinjlim\mathscr{F}_i(U\cup V).$ Since the restriction map $\varinjlim\mathscr{F}_i(U\cup V)\to\varinjlim\mathscr{F}_i(U)$ commutes with the restriction map $\mathscr{F}_j(U\cup V)\to\mathscr{F}_j(U)$ we see that $r|_U=s$ since $r&39;|_U=s&39;.$ Similarly $r|_V=t.$

Now since $X$ is Noetherian, any open cover $\{U_\alpha\}$ of an open set $U\subseteq X$ has a finite subcover $\{U_1,\ldots,U_r\}.$ Thus we may glue sections of arbitrary open covers iteratively via the pairwise gluing outlined above. Hence the direct limit presheaf is in fact a sheaf.

1.12. Inverse limits. Suppose that $\{U_\alpha\}$ is an open cover of $U\subseteq X$ and $s_\alpha\in\varprojlim\mathscr{F}_i(U_\alpha)$ such that the $s_\alpha$ are agree on intersections. Then for each projection map $\pi_k:\varprojlim\mathscr{F}_i(U)\to\mathscr{F}_k(U)$ the sections $\pi_k(s_\alpha)\in\mathscr{F}_k(U_\alpha)$ agree on intersections as well. Thus for each $k$ the sections $\pi_k(s_\alpha)$ glue to a section $s^k\in\mathscr{F}_k(U).$ We may explicitly write $\varprojlim\mathscr{F}_i(U):= \left\{(a_i)\in\prod\mathscr{F}_i(U) : \varphi_{ij}(U)(a_i)=a_j \text{ for all } i,j\right\}$ with $\varphi_{ij}:\mathscr{F}_i\to\mathscr{F}_j$ the maps of the directed system of sheaves. Thus, since $\varphi_{ij}(s^i)=s^j$ for all $i,j,$ above we in fact constructed an element $s=(s^k)\in\varprojlim\mathscr{F}_i(U).$ Moreover, $s|_{U_\alpha}=s_\alpha$ as desired. Thus $\varprojlim\mathscr{F}_i$ is a sheaf.

This object clearly satisfies the properties of an inversel limit in the category of sheaves since it satisfies the inverse limit property locally for each open set.

1.13. Espace Étale of a Presheaf. See the corresponding Eisenbud exercise I-8..

1.14. Support. Suppose $Q\in U\setminus\text{Supp}(s).$ Then $s_Q=0$ so there exists an open neighborhood $V\subseteq U$ of $Q$ such that $s|_{V}=0.$ So $V\subseteq U\setminus\text{Supp}(s).$ Thus $\text{Supp}(s)$ is open.

Let us also describe an example sheaf $\mathscr{F}$ such that $\text{Supp}(\mathscr{F})$ is not a closed set. This example was given by a fellow grad student Ahmad Mokhtar during a Hartshorne seminar at SFU. Let $X=\mathbb{R}$ and for $r\in\mathbb{R}$ define $\mathscr{Z}_{r}$ to be the skyscraper sheaf of $\mathbb{Z}$ over the point $r.$ Then take $\mathscr{F}=\bigoplus_{n\in\mathbb{N}}\mathscr{Z}_{1/n}$ where the infinite direct sum sheaf is defined to be the sheafification of the presheaf infinite direct sum $\mathscr{P}.$ Observe that on the presheaf infinite direct product $\mathscr{P}$ any section section $s\in\mathscr{P}(U)$ has a finite number of points in its support. Therefore for any $r\ne 1/n$ for $n\in\mathbb{N}$ the stalk $\mathscr{P}_r$ is $0$ since given any section $s\in\mathscr{P}(U)$ with $U$ a neighborhood of $r,$ we may choose a smaller open neighborhood $V\subset U$ such that the support of $s$ does not lie in $V$ and hence $s|_V=0.$ Thus $\mathscr{P}_r=0$ if $r\not\in\{1/n \, : \, n\in\mathbb{N}\}$ and $\mathscr{P}_{1/n}=\mathbb{Z}$ for each $n\in\mathbb{N}.$ Since sheafification does not change the stalks, the same holds for $\mathscr{F}.$ So $\text{Supp}(\mathscr{F})=\{1/n \, : \, n\in\mathbb{N}\}$ which is not closed in $X=\mathbb{R}.$

1.15. Observe that if $\varphi,\psi:\mathscr{F}|_U\to\mathscr{G}|_U$ is a morphism then so are $(\varphi+\psi),(-\varphi):\mathscr{F}|_U\to\mathscr{G}|_U$ defined locally for each open $U\subset V$ by $(\varphi+\psi)(V) =\varphi(V)+\psi(V)$ and $(-\varphi)(V)=-\varphi(U),$ respectively. Thus $\text{Hom}(\mathscr{F}|_U,\mathscr{G}|_U)$ is an abelian group with addition defined as above and the zero morphism as the identity.

1.16. Flasque sheaves.

1.16.a. Let us first recall that if $X$ is irreducible then any nonempty subset $U\subseteq X$ is connected. To see this, observe that if $U=U_1\cup U_2$ with $U_1,U_2$ nonempty proper open subsets, then $X\setminus(U_1\cap U_2)=(X\setminus U_1)\cup (X\setminus U_2).$ Therefore $U_1\cap U_2\ne\emptyset$ since $X\setminus U_1$ and $X\setminus U_2$ are nonempty proper closed subsets of $X.$

Let $\mathscr{F}$ be the constant sheaf associated to a group $A$ on an irreducible space $X.$ Observe that since each nonempty open $U\subseteq X$ is connected, $\mathscr{F}(U)=A.$ Moreover, all restriction maps must be the identity (unless restricting to $\mathscr{F}(\emptyset)=0$ ). Hence restriction maps are surjective and $\mathscr{F}$ is flasque.

Notation. Let us use $\varphi$ to denote the morphism $\mathscr{F}&39;\to\mathscr{F}$ and $\psi$ to denote the morphism $\mathscr{F}\to\mathscr{F}&39;&39;.$

1.16.b. By Exercise 1.1.8 it suffices to show that the map $\psi(U):\mathscr{F}(U)\to\mathscr{F}&39;&39;(U)$ is surjective.

Claim 1. Let $s\in\mathscr{F}&39;&39;(U).$ Suppose that $U_1,U_2\subseteq U$ are open such that there exist $t_1\in\mathscr{F}(U_1)$ and $t_2\in\mathscr{F}(U_2)$ with $\psi(U_1)(t_1)=s|_{U_1}$ and $\psi(U_2)(t_2)=s|_{U_2}.$ Then there exists $t\in\mathscr{F}(U_1\cup U_2)$ such that $\psi(U_1\cup U_2)(t)=s|_{U_1\cup U_2}.$

Let $U_1,U_2\subseteq U$ be open sets such that there exist $t_1\in\mathscr{F}(U_1)$ and $t_2\in\mathscr{F}(U_2)$ with $\psi(U_1)(t_1)=s|_{U_1}$ and $\psi(U_2)(t_2)=s|_{U_2}.$ Then $t_1|_{U_1\cap U_2}-t_2|_{U_1\cap U_2}=u\in\text{ker}(\psi(U_1\cap U_2))= \text{im}(\varphi(U_1\cap U_2)).$ Since $\mathscr{F}&39;$ is flasque, there exists $u_1\in\mathscr{F}&39;(U_1)$ such that $u_1|_{U_1\cap U_2}=u.$ Thus $\psi(U_1)(t_1-\varphi(u_1))=\psi(U_1)(t_1)$ and $(t_1-\varphi(u_1))|_{U_1\cap U_2}=t_2|_{U_1\cap U_2}.$ So by the sheaf axiom $t_1-\varphi(u_1)$ and $t_2$ lift to a section $t\in\mathscr{F}(U_1\cup U_2)$ such that $\psi(U_1\cup U_2)(t)=s|_{U_1\cup U_2}.$

Claim 2. The map $\psi(U)$ is surjective.

Let $s\in\mathscr{F}&39;&39;(U).$ Since $\psi$ is surjective there exists an open cover $\{V_\alpha\}_{\alpha\in A}$ of $U$ and $t_\alpha\in\mathscr{F}(V_\alpha)$ such that $\psi(V_\alpha)(t_\alpha)=s|_{V_\alpha}$ (see Corollary 3 below).

For any subset $B\subseteq A$ define $U_B=\bigcup_{\beta\in B}V_\beta.$ Auppose that $\{V_\alpha\}$ is a minimal cover, i.e., $U_B\ne U$ if $B\ne A.$

Let us define the set $S = \left\{(t,B) \, : \, B\subseteq A, \, t\in\mathscr{F}(U_B), \psi(U_B)(t)=s|_{U_B}\right\}.$ Observe that $S$ is nonempty since $(t_\alpha,\{\alpha\})\in S$ for each $\alpha\in A.$ Give $S$ a partial order by $(u_1,B_1)\le (u_2,B_2)$ if $B_1\subseteq B_2$ and $u_2|_{U_{B_1}}=u_1.$

Let $\{(u_i, B_i)\}_{i\in I}$ be a chain in $S,$ i.e., $I$ is some totally ordered set and $(u_i,B_i)\le (u_j,B_j)$ whenever $i\le j.$ We will show that every such chain has an upper bound, with the objective of applying Zorn’s lemma to construct a maximal element of $S.$

Observe that $V=\bigcup_{i\in I}U_{B_i}$ is an open subset of $V$ covered by $\{U_{B_i}\}_{i\in I}.$ Moreover, $u_i|_{U_i\cap U_j}=u_j|_{U_i\cap U_j}$ for all $i,j\in I.$ Thus by the sheaf axiom, there exists $u\in\mathscr{F}(V)$ such that $u|_{U_{B_i}}=u_i$ for all $i\in I.$ The element $(u,\bigcup_{i\in I}B_i)$ is an upper bound for the chain $\{(u_i, B_i)\}$ in $S.$

By Zorn’s lemma, there exists a maximal element $(u,B)$ in $S.$ Suppose that $B\ne A,$ i.e., $U_B\ne U.$ Then pick $\alpha$ such that $V_\alpha\not\subseteq U_B$ and apply Claim 1 to obtain a section $u&39;\in\mathscr{F}(U_B\cup V_\alpha)$ such that $\psi(U_B\cup V_\alpha)(u&39;)=s|_{U_B\cup V_\alpha}.$ Then $(u&39;,B\cup\{\alpha\})\not\le (u,B),$ a contradiciton. So it must be that $B=A$ and $u\in\mathscr{F}(U)$ such that $\psi(U)(u)=s.$

1.16.c. Let $V\subseteq U\subseteq X$ be open subsets. Let $s\in\mathscr{F}&39;&39;(V).$ Let us use $\psi$ to denote the surjective morphism $\mathscr{F}\to\mathscr{F}&39;&39;.$ By part (b) the map $\psi(V):\mathscr{F}(V)\to\mathscr{F}&39;&39;(V)$ is surjective. Since $\mathscr{F}$ is flasque, the restriction map $\mathscr{F}(U)\to\mathscr{F}(V)$ is surjective. Thus there exists $t\in\mathscr{F}(U)$ such that $\psi((t|_{V})=s.$ But $\psi(t|_{V})=\psi(t)|_{V}$ and thus $\psi(t)\in\mathscr{F}&39;&39;(U)$ is such that $\psi(t)|_V=s.$ Hence the restriction map $\mathscr{F}&39;&39;(U)\to\mathscr{F}&39;&39;(V)$ is surjective.

1.16.d. Suppose $V\subseteq U\subseteq Y$ are open sets. Then since $\mathscr{F}$ is flasque and $f^{-1}(V)\subseteq f^{-1}(U)\subseteq X$ are open, the restriction map $\mathscr{F}(f^{-1}(U))\to\mathscr{F}(f^{-1}(V))$ is surjective. Thus the restriction map $f_*\mathscr{F}(U)=\mathscr{F}(f^{-1}(U))\to\mathscr{F}(f^{-1}(V)) =f_*\mathscr{F}(V).$

1.16.e. By exercise I-8 in Eisenbud we may view $\mathscr{F}$ as the sheaf of germs of continuous sections of the projection map $\pi:\overline{\mathscr{F}}\to X$ where $\overline{\mathscr{F}}=\bigsqcup_{x\in X}\mathscr{F}_x=\{(x,s_x):x\in X, s_x\in\mathscr{F}_x\}.$ Thus the sheaf of discontinuous functions defined above simply removes the restriction that sections be continuous. Thus it is easy to see that $\mathscr{G}$ is a sheaf and that there is a natural inclusion of $\mathscr{F}$ in $\mathscr{G}.$

It remains to see that $\mathscr{G}$ is flasque. Suppose that $V\subseteq U\subseteq X$ are open and $s\in\mathscr{G}(V).$ Then we may define $t\in\mathscr{G}(U)$ by $t(P)=s(P)$ if $P\in U$ and $t(P)=0$ otherwise. Since $t|_{V}=s,$ this shows that the restriction map is surjective.

1.17. Suppose $Q\in\overline{\{P\}}.$ Suppose that $s_1\in i_P(U_1)$ and $s_2\in i_P(U_2)$ for $U_1,U_2$ containing $Q,$ then $Q\in U_1\cap U_2$ implies that $P\in U_1\cap U_2$ by our choice of $Q.$ Thus we may consider $s_1,s_2$ as elements of $A$ and $s_1=s_2$ in $A$ if and only if $s_1|_{U_1\cap U_2}=s_2|_{U_1\cap U_2}.$ So $\mathscr{F}_Q=A.$ If, on the other hand, $Q\not\in\overline{\{P\}},$ then there exists an open neighborood $U$ of $Q$ with $P\not\in U.$ Thus $\mathscr{F}_Q=0.$

By definition of a pushforward, this is exactly the sheaf given by taking the pushforward $i_*(A)$ under the given inclusion map $i.$

1.18. Adjoint Property of $f^{-1}$ .

Note. Explicit representation of direct limits. We use the following construction of direct limits: if $A_\alpha$ is a directed system of groups with maps $\phi_{\alpha,\beta}:A_{\alpha}\to A_{\beta},$ then $\varinjlim A_\alpha=\{(s,A_\alpha) : s\in A_\alpha\}/\sim$ where $(s,A_\alpha)\sim (t,A_\beta)$ if there exists $\gamma$ such that $\phi_{\alpha,\gamma}(s)=\phi_{\beta,\gamma}(t).$ In this way we may also define addition $[(s,A_{\alpha})]+[(t,A_{\beta})]=[(\phi_{\alpha,\gamma}(s)+\phi_{\beta,\gamma}(t), A_{\gamma}]$ where $\gamma\ge \alpha,\beta.$

For a given topological space $X$ let $\text{PSh}_X$ be the category of presheaves on $X$ and let $\text{Sh}_X$ be the category of sheaves on $X.$

We will show below that $f^{-1}:\text{PSh}_Y\to\text{PSh}_X$ without sheafification is a left adjoint to $f_*:\text{PSh}_X\to\text{PSh}_Y.$ Thus for every $\mathscr{F}\in\text{PSh}_X$ and every $\mathscr{G}\in\text{PSh}_Y$ we will have a bijection $\text{Hom}_{PSh}(f^{-1}\mathscr{G},\mathscr{F}) \cong\text{Hom}_{PSh}(\mathscr{G},f_*\mathscr{F}).$ Recall that if $\mathscr{P}$ is a presheaf on $X$ and $\mathscr{G}$ is a sheaf on $X,$ then we also have a bijection $\text{Hom}_{PSh}(\mathscr{P},\mathscr{G}) \cong\text{Hom}_{Sh}(\mathscr{P}^+,\mathscr{G}).$ Thus for any $\mathscr{F}\in\text{Sh}_X$ and $\mathscr{G}\in\text{Sh}_Y,$ the result on presheaves gives us that, when we consider $f^{-1}$ with sheafification, we have an adjoint bijection $\text{Hom}_{Sh}(f^{-1}\mathscr{G},\mathscr{F}) \cong \text{Hom}_{Sh}(\mathscr{G},f_*\mathscr{F}).$ Therefore it in fact suffices to show that $f^{-1}:\text{PSh}_Y\to\text{PSh}_X$ without sheafification is a left adjoint to $f_*:\text{PSh}_X\to\text{PSh}_Y.$

Recall that to show $f^{-1}$ is a left adjoint to $f_*$ it is equivalent to define counit and unit natural transformations $\epsilon:f^{-1}f_*\to1_{\text{PSh}_X}$ and $\eta:1_{\text{PSh}_Y} \to f_*f^{-1},$ and show that for any $\mathscr{F}\in\text{PSh}_X$ and $\mathscr{G}\in\text{PSh}_Y$ we have $f_*\epsilon_{\mathscr{F}}\circ\eta_{f_*\mathscr{F}}=1_{f_*\mathscr{F}}$ and $\epsilon_{f^{-1}\mathscr{G}}\circ f^{-1}\eta_{\mathscr{G}}=1_{f^{-1}\mathscr{G}}.$

Observe that $\begin{aligned}[t] f^{-1}f_*\mathscr{F}(U) &=\varinjlim_{V\supset f(U)}f_*\mathscr{F}(U)\\ &=\varinjlim_{V\supset f(U)}\mathscr{F}(f^{-1}(V)). \end{aligned}$ Therefore we have a natural map $f^{-1}f_*\mathscr{F}(U)\to \mathscr{F}(U)$ defined by $[(s,\mathscr{F}(f^{-1}(V)))]\mapsto s|_{U}.$ This collection of maps defines a map of presheaves, as restriction on each presheaf commutes. For a given presheaf $\mathscr{F}$ on $X$ we will denote this map as $\epsilon_{\mathscr{F}}:f^{-1}f_*\mathscr{F}\to\mathscr{F}.$ The collection of presheaf maps will define our counit natural transformation $\epsilon.$

Next, we explcitly write out $\begin{aligned}[t] f_*f^{-1}\mathscr{G}(U)&=f^{-1}\mathscr{G}(f^{-1}(U))\\ &=\varinjlim_{V\supset f(f^{-1}(U))}\mathscr{G}(V). \end{aligned}$ Since $U\supset f(f^{-1}(U))$ we have the natural map $\mathscr{G}(U)\to f_*f^{-1}\mathscr{G}(U)$ defined by $s\mapsto [(s,\mathscr{G}(U))].$ Again this commutes with restriction and hence defines a map of presheaves. Again for a given (pre)sheaf $\mathscr{G}$ on $Y$ we will denote the this map as $\eta_{\mathscr{G}}:\mathscr{G}\to f_*f^{-1}\mathscr{G}$ and this collection of presheaf maps will define our unit natural transformation $\eta.$

It is routine to verify that $\epsilon,\eta$ defined above are natural transformations. Let us work this out explicitly for $\epsilon.$ If $\varphi:\mathscr{F}_1\to\mathscr{F}_2$ is a map of presheaves on $X$ then we need to show that the following diagram commutes: $\xymatrix{ f^{-1}f_*\mathscr{F}_1(U) \ar[r]^{\epsilon_{\mathscr{F}_1}(U)} \ar[d]^{(f^{-1}f_*\varphi)(U)} & \mathscr{F}_1(U) \ar[d]^{\varphi(U)}\\ f^{-1}f_*\mathscr{F}_2(U) \ar[r]^{\epsilon_{\mathscr{F}_2}(U)} & \mathscr{F}_2(U) }$ Tracing a particular element $[(s,f^{-1}(V))]\in f^{-1}f_*\mathscr{F}_1(U),$ with $U\subseteq X$ open, through the diagram we find $\xymatrix{ [(s,f^{-1}(V)] \ar[r] \ar[d] & s|_U \ar[d]\\ [(\varphi(f^{-1}(V))(s),f^{-1}(V))] \ar[r] & \varphi(U)(s|_U) }$ since $\varphi(f^{-1}(V))(s)|_U=\varphi(U)(s|_U)$ by the commutativity of presheaf maps with presheaf restriction. A similar diagram chase shows that $\eta$ is also a natural transformation.

It now remains to show that for any presheaf $\mathscr{F}$ on $X$ and any presheaf $\mathscr{G}$ on $Y$ we have $f_*\epsilon_{\mathscr{F}}\circ\eta_{f_*\mathscr{F}}=1_{f_*\mathscr{F}}$ and $\epsilon_{f^{-1}\mathscr{G}}\circ f^{-1}\eta_{\mathscr{G}}=1_{f^{-1}\mathscr{G}}.$ Again diagram chasing will suffice. We have a sequence of maps $f_*\mathscr{F}\overset{\eta_{f_*\mathscr{F}}}{\to} f_*f^{-1}f_*\mathscr{F}\overset{f_*\epsilon_{\mathscr{F}}}{\to} f_*\mathscr{F}.$ Tracing through a particular element $s\in f_*\mathscr{F}(U)=\mathscr{F}(f^{-1}(U)),$ with $U\subseteq Y$ open, we find $\begin{aligned}[t] (f_*\epsilon_{\mathscr{F}})(\eta_{f_*\mathscr{F}}(s)) &= (f_*\epsilon_{\mathscr{F}})\left([(s,U)]\right)\\ &= s|_U\\ &= s. \end{aligned}$ Note that the maps above should technically have $``(U)"$ in front of them to indicate that they are the local maps, but this was omitted to attempt to make things less cluttered. The above follows from an explicit description of $f_*f^{-1}f_*\mathscr{F}$ as $\begin{aligned}[t] f_*f^{-1}f_*\mathscr{F}(U) &= f^{-1}f_*\mathscr{F}(f^{-1}(U))\\ &= \varinjlim_{V\supseteq f(f^{-1}(U))}f_*\mathscr{F}(V)\\ &= \varinjlim_{f^{-1}(V)\supseteq f^{-1}(f(f^{-1}(U)))}\mathscr{F}(f^{-1}(V))\\ &= \varinjlim_{f^{-1}(V)\supseteq f^{-1}(U)}\mathscr{F}(f^{-1}(V)). \end{aligned}$ Next, we have the sequence of maps $f^{-1}\mathscr{G}\overset{f^{-1}\eta_{\mathscr{G}}}{\to} f^{-1}f_*f^{-1}\mathscr{G}\overset{\epsilon_{f^{-1}\mathscr{G}}}{\to} f^{-1}\mathscr{G}.$ Again a diagram chase shows us that for $[(s,V)]\in f^{-1}\mathscr{G}(U)=\varinjlim_{V\supseteq f(U)}\mathscr{G}(V),$ with $U\subseteq X$ open, we have $\begin{aligned}[t] \epsilon_{f^{-1}\mathscr{G}}((f^{-1}\eta_{\mathscr{G}})([(s,V)])) &= \epsilon_{f^{-1}\mathscr{G}}([([(s,V)], U])\\ &= [(s,V)]. \end{aligned}$ The above follows from an explicit descriptin of the object $f^{-1}f_*f^{-1}\mathscr{G}(U)$ as $\begin{aligned}[t] f^{-1}f_*f^{-1}\mathscr{G}(U) &= \varinjlim_{V\supset f(U)}\left(f_*f^{-1}\mathscr{G}(V)\right)\\ &= \varinjlim_{V\supseteq f(U)}\left( \varinjlim_{W\supseteq f(f^{-1}(V))}\mathscr{G}(W)\right). \end{aligned}$

1.19. Extending a Sheaf by Zero.

1.19.a. Let $P\in\mathbb{Z}.$ Observe that the open subsets of $Z$ containing $P$ are exactly the subsets of the form $i^{-1}(V)=V\cap Z$ where $U$ is an open subset of $X.$ Thus $i_*(\mathscr{F})_P=\mathscr{F}_P.$

Suppose that $P\not\in\mathbb{Z}.$ Let $s\in i_*(\mathscr{F})(V)$ for some $U$ containing $P.$ Then $V\cap U$ is also an open set containing $P$ and $s|_{V\cap U}=0.$ Thus $i_*(\mathscr{F})_P=0.$

1.19.b. Observe that the open subsets of $U$ are exactly the open subsets of $X$ contained in $U.$ Thus $j_!(\mathscr{F})_P=\mathscr{F}_P$ for $P\in U.$ If $P\not\in U$ then no open subset containing $P$ is contained in $U.$ Thus $j_!(\mathscr{F})_P=0.$

1.19.c. Observe by parts (a) and (b) for each $P\in X$ we have an exact sequence of stalks $0\to j!(\mathscr{F}|_{U})_P\to\mathscr{F}_P\to i_*(\mathscr{F}|_U)_P\to 0.$ Thus the result follows by the fact that exactness on stalks is equivalent exactness of sheaves.

Suppose that $\mathscr{G}$ is another sheaf on $X$ such that $G|_{U}=\mathscr{F},$ and such that $G_P=\mathscr{F}_P$ if $P\in U$ and $G_P=0$ otherwise. Let $\mathscr{P}$ be the presheaf given by $V\mapsto\mathscr{F}(V)$ for $V\subseteq U$ and $V\mapsto 0$ otherwise. Then clearly there is a map of presheaves from $\mathscr{P}\to\mathscr{G}$ that induces an isomorphism on the stalks. By sheafification we have a map of sheaves from $j_!(\mathscr{F})=\mathscr{P}^+\to\mathscr{G}.$ This sheaf map also induces an isomorphism on the stalks $j_!(\mathscr{F})_P\cong\mathscr{G}_P$ for each $P\in X.$ Thus $j_!(\mathscr{F})\cong\mathscr{G}.$

1.20. Subsheaf with Supports

1.20.a. Suppose we have a cover $\{V_{\alpha}\}$ of an open set $V\subseteq X$ and $s_{\alpha}\in\mathscr{H}_Z^0(\mathscr{F})(V_{\alpha})$ such that $s_{\alpha}|_{V_\alpha\cap V_\beta}=s_{\beta}|_{V_\alpha\cap V_\beta}$ for all $\alpha,\beta.$ Then since $\mathscr{F}$ is a sheaf there exists an $s\in\mathscr{F}(V)$ such that $s|_{V_\alpha}=s_\alpha$ for all $\alpha.$ In particular, $\text{Supp}(s)=\bigcup \text{Supp}(s_{\alpha})\subseteq Z.$ So $s\in\mathscr{H}_Z^0(\mathscr{F})(V).$

1.20.b. Let $s\in\mathscr{F}(V)$ for some open $V\subseteq X.$ Observe that $s\mapsto s|_{j^{-1}(V)}\in j_*(\mathscr{F}|_{U}).$ Since $j^{-1}(V)=V\cap U,$ $s\mapsto 0$ if and only if $\text{Supp}(s)\subseteq V\setminus U=V\cap Z,$ that is, if and only if $s\in\mathscr{H}_Z^0(\mathscr{F})(V).$

If $\mathscr{F}$ is flasque, then the restricton map $\mathscr{F}(V)\mapsto \mathscr{F}(V\cap U)=j_*(\mathscr{F}|_U)(V)$ is surjective. Thus the map $\mathscr{F}\to j_*(\mathscr{F}|_U)$ is surjective.

1.21. Some Examples of Sheaves on Varieties.

1.21.a. By a similar argument to exercise 1.20.a it is easy to see that $\mathscr{I}_Y$ is a sheaf: sections in $\mathscr{O}_X$ with support in $Y$ always glue to sections with support in $Y.$

1.21.b. Suppose that $X\subset\mathbb{P}^n$ is a quasiprojective variety and $Y$ a subvariety. Let $i:Y\to X$ be the inclusion map. Observe that we have a natural map of sheaves from $\mathscr{O}_X\to i_*(\mathscr{O}_Y)$ given by restriction of functions. Moreover, the kernel of this map is just those functions that vanish on $Y,$ that is, the sheaf $\mathscr{I}_Y.$ If we now show that the map is surjectvie exercise 1.7.a gives us that $i_*(\mathscr{O}_Y)\cong\mathscr{O}_X/\mathscr{I}_Y.$

Suppose that $s\in\mathscr{O}_Y(U)$ for some open set $U\subseteq Y.$ Note that $U=V\cap Y$ for some open $V\subset X.$ Then for each $P\in U$ there exists an open set $U_P\subseteq U$ such that $s|_{U_P}=g_P/h_P$ for some $h_p,g_P\in k[x_0,\ldots,x_n].$ By possibly shrinking $U_P$ we may assume that $U_P\subseteq (U\setminus V(h_P)).$ We may also assume that $U_P=V_p\cap Y$ for some $V_p\subseteq (X\setminus V(h_P))$ open in $X.$ Therefore $g_P/h_P$ is a regular function on $V_P$ , i.e. an element of $\mathscr{O}_X(V_p),$ and $g_P/h_P\mapsto s|_{U_P}$ under the map $\mathscr{O}_X(V_P)\to i_*(\mathscr{O}_Y)=\mathscr{O}_Y(U_P).$ Thus the map $\mathscr{O}_X\to i_*(\mathscr{O}_Y)$ is surjective by exercise 1.3a.

1.21.c. Suppose that $P\in Y.$ Let $s\in\mathscr{I}_Y(V)$ for some open $V$ containing $P.$ Note that $\text{Supp}(s)$ is a closed subset of $V$ disjoint from $Y,$ and thus $W=V\setminus\text{Supp}(s)$ is an open set containing $P$ such that $s|_W=0.$ Thus $\mathscr{I}_{Y,P}=0.$ Conversely, if $P\not\in Y,$ then a similar argument to exercise 1.19.b shows that $\mathscr{I}_{Y,P}=\mathscr{O}_{P}.$ Thus the given sequence is exact on the stalks and hence exact. Note that the sequence of global sections is not exact since global sections of $\mathscr{O}_{\mathbb{P}^1}$ are just the constants $k,$ and $\mathscr{O}_P=\mathscr{O}_Q=k.$ There is no surjective map $k\to k^2.$

1.21.d. By the definition of $\mathscr{O}_X$ we have $\mathscr{O}_X(U)\subseteq K$ for each $U.$ Therefore there is an injection $\varphi:\mathscr{O}_X\hookrightarrow\mathscr{K}$ given locally by sending $s\in\mathscr{O}_X(U)$ to the constant map $U\mapsto \{s\}\subseteq K$ in $\mathscr{K}(U).$ Therefore we have an exact sequence $0\to\mathscr{O}_X\to\mathscr{K}\to\mathscr{K}/\mathscr{O}_X\to 0.$ It suffices to show that for any $P\in X$ we have $\mathscr{K}_P= K.$ Suppose that $U$ is an open neighborhood of $P$ and $s\in\mathscr{K}(U).$ By continuity, $V=s^{-1}(s(P))$ is an open neighborhood of $P.$ Therefore $s|_V$ is constant. Hence $\mathscr{K}_P=K.$

1.21.e. Observe that the map $\mathscr{K}\to\mathscr{K}/\mathscr{O}_X\cong \sum_{P\in X}i_P(I_P)$ is given by $s\mapsto\sum_{P\in X} s(P)+\mathscr{O}_P.$ Therefore $s\in\Gamma(X,\mathscr{K})$ is such that $s\mapsto 0$ if and only if $s\in\mathscr{O}_P$ for all $P\in X,$ i.e., if and only if $s\in\Gamma(X,\mathscr{O}_X).$ Thus the sequence of global sections is exact at $\Gamma(X,\mathscr{K}).$

It remains to show that the map $\Gamma(X,\mathscr{K})\to\Gamma(X,\mathscr{K}/\mathscr{O}_X)$ is surjective. By part (d) we have that $\Gamma(X,\mathscr{K}/\mathscr{O}_X)=\sum_{P\in X} K/\mathscr{O}_P.$ Thus a nonzero global section $s\in\Gamma(X,\mathscr{K}/\mathscr{O}_X)$ is a tuple $(s_{P})_{P\in X}$ such that $s_{P}\in\mathscr{O}_P$ for all but finitely many points $P_1,\ldots,P_r.$ Thus in order to show that the map $\Gamma(X,\mathscr{K})\to\Gamma(X,\mathscr{K}/\mathscr{O}_X)$ is surjective, it suffices to show that it is surjective onto each term $K/\mathscr{O}_P$ of the direct product. That is, we must show that for any function $f\in K$ and every $P\in X,$ there exists a function $g\in K$ such that $f-g\in\mathscr{O}_P$ and $g\in\mathscr{O}_Q$ for all $Q\in X\setminus\{P\}.$

By a linear automorphism of $\mathbb{P}^1$ we may ensure that $P=[0,1]$ (see my solutions to Fulton’s curve book). Let us consider just the affine patch $\mathbb{A}^1$ given by $y\ne 0$ in $\mathbb{P}^1,$ recalling that $K\cong k(x)=k(\mathbb{A}^1).$ Note that $P=0\in\mathbb{A}^1.$ Write $f=\frac{a(x)}{x^n b(x)}$ where $x\nmid a$ and $x\nmid b.$ If $n\le 0$ then $f$ is regular at $P$ and therefore any constant function $f&39;\in k$ satisfies our requirements. Suppose that $n>0.$ Since $x\nmid b$ we may construct a formal power series inverse to $b$ in which we will call $B^{-1}\in k[[x]].$ Let us take the first $n$ terms of the power series $aB^{-1}$ and call this $p(x).$ We then find that $x^n\mid (a(x)-p(x)b(x)).$ Thus if we define $f&39;=\frac{p(x)}{x^n}$ then $f-f&39;=\frac{a(x)-p(x)b(x)}{x^nb(x)}$ is regular at $P.$ Moreover, $f&39;$ is clearly regular at all points other than zero, including infinity.

Accreditation. I was slightly lazy with part (e) and ended up finding the idea to use a “power series inverse” style polynomial in the solution to this exercise by Cutrone and Mashburn.