# Problem Sets

## Problem Set 1

All problems are sourced from Hartshorne.

## Chapter 2, Section 1

List of useful facts used:

• Morphisms of sheaves commute with restrictions: if $$\phi: {\mathcal{F}}\to {\mathcal{G}}$$ then for any $$s\in {\mathcal{F}}(U)$$ and $$V \subseteq U$$, $$\mathop{\mathrm{Res}}(U, V)(\phi(s)) = \phi(\mathop{\mathrm{Res}}(U, V)(s))$$.
• $$\phi$$ is an isomorphism iff $$\phi_p$$ are all isomorphisms.
• Elements of stalks $${\mathcal{F}}_p:$$ equivalence classes $$[U, s\in {\mathcal{F}}(U)]$$.
• The induced map on stalks: $$\phi_p([U, s]) \coloneqq[U, \phi(U)(s)]$$.
• A surjection of sheaves need not induce a surjection on sections.
• The colimit diagram:

• Colimits are initial co-cones, where $$I$$ is initial if $$I\to X$$ for any $$X$$. AKA direct limits.

• Filtered colimits commute with finite limits.

• In particular, monomorphisms are pullbacks, so finite limits, and stalks are filtered colimits. So injections of sheaves induce injections on stalks.

Recommended problems:

• 1.1
• 1.2
• 1.3
• 1.4
• 1.5

Let $$A$$ be an abelian group, and define the constant presheaf associated to $$A$$ on the topological space $$X$$ to be the presheaf $$U \mapsto A$$ for all $$U \neq \emptyset$$, with restriction maps the identity.

Show that the constant sheaf $${\mathcal{A}}$$ defined in Hartshorne is the sheafification of this presheaf.

Let $$X\in {\mathsf{Top}}$$ be a space. Recapping the definitions, define the constant presheaf as \begin{align*} \underline{A}^{\mathsf{pre}}(U) \coloneqq \begin{cases} A & U\neq \emptyset \\ 0 & \text{else}. \end{cases} \quad \operatorname{res}^1(U, V) \coloneqq \begin{cases} \operatorname{id}_A & U\neq \emptyset \\ 0 & \text{else}. \end{cases} .\end{align*}

Then define the constant sheaf as \begin{align*} \underline{A}(U) \coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(U, A)\quad \operatorname{res}^2(U, V)(f) \coloneqq{ \left.{{f}} \right|_{{V}} } .\end{align*}

We’re then tasked with finding a morphism of sheaves \begin{align*} \Psi: (\underline{A}^{\mathsf{pre}})^+ \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Top}}}({-}, A) ,\end{align*} which we’ll also want to have an inverse morphism and this define an isomorphism in $${\mathsf{Sh}}(X)$$.

We’ll use the implicitly stated fact in Hartshorne that $$\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(U, A) = A^{\oplus n}$$ where $$n \coloneqq# \pi_0(X)$$ is the number of connected components of $$U$$. Suppose first that $$n=1$$, so $$X$$ is connected, and define the following morphism of groups: \begin{align*} \Psi_U: (\underline{A}^{\mathsf{pre}})(U) = A &\longrightarrow\mathop{\mathrm{Hom}}_{\mathsf{Top}}(U, A)\\ a_0 &\mapsto \left\{ { \begin{aligned} \varphi_{a_0}: U \to A \\ x \mapsto a_0, \end{aligned} } \right. \end{align*} which maps a group element $$a_0$$ to the constant function on $$U$$ sending every point to $$a_0 \in A$$. The claim is that the following diagram commutes in the category $$\underset{ \mathsf{pre} } {\mathsf{Sh} }(X)$$ (in both directions) for all $$U$$ and $$V$$:

Here we’ve specified simultaneously what $$\Psi$$ and $$\Psi^{-1}$$ prescribe on opens $$U, V$$, and abuse notation slightly by writing $$\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}({-}, A)$$ for the sheaf it represents and its underlying presheaf.

• That this commutes follows readily, since running the diagram counterclockwise yields $$\operatorname{res}^1(U, V) = \operatorname{id}_A$$, so the composition \begin{align*} (A \xrightarrow{\operatorname{res}^1(U, V)} A \xrightarrow{\Psi_V} \mathop{\mathrm{Hom}}(V, A)) = (A \xrightarrow{\Psi_V} \mathop{\mathrm{Hom}}(V, A)) \end{align*} sends an element $$a_0\in A$$ to the constant function $$\varphi_{a_0, V}: V\to A$$. Running the diagram clockwise yields \begin{align*} (A \xrightarrow{\Psi_U} \mathop{\mathrm{Hom}}(U, A) \xrightarrow{\operatorname{res}^2(U, V)} \mathop{\mathrm{Hom}}(V, A)) ,\end{align*} which sends $$a_0$$ to the constant function $$\varphi_{a_0, U}: U\to A$$ sending everything to $$a_0$$, which then gets sent to $${ \left.{{\varphi_{a_0, U}}} \right|_{{V}} }: V\to A$$ sending everything to $$a$$. Since $${ \left.{{\varphi_{a_0}}} \right|_{{V}} }(x) = \varphi_{a_0, V}(x) = a$$ for every $$x\in U$$, these functions are equal.

• That the reverse maps $$\Psi_U^{-1}$$ are well-defined follows from the fact that $$U$$ is connected: the continuous image of a connected set is connected. Since $$A$$ is given the discrete topology, any subset with 2 or more elements in disconnected, so each function $$f\in \mathop{\mathrm{Hom}}(U, A)$$ is necessarily a constant function and $$f(U) = \left\{{a}\right\}$$ is a singleton.

• $$\Psi_U, \Psi_U^{-1}$$ clearly compose to the identity in either order, so $$\Psi_U$$ defines an isomorphism of abelian groups.

As a consequence, we get a well-defined morphism of presheaves $$\underline{A}^{\mathsf{pre}}({-}) \to { \left.{{ \mathop{\mathrm{Hom}}({-}, A)}} \right|_{{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}} }$$, and by the sheafification adjunction we can lift this to a morphism of sheaves: \begin{align*} \adjunction{{\mathcal{F}}\mapsto {\mathcal{F}}^+ }{{\mathcal{G}}\mapsto { \left.{{{\mathcal{F}}}} \right|_{{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}} } }{ \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)}{{\mathsf{Sh}}(X)} ,\end{align*} which reads \begin{align*} \mathop{\mathrm{Hom}}_{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}({\mathcal{F}}, { \left.{{{\mathcal{G}}}} \right|_{{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}} }) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Sh}}}({\mathcal{F}}^+, {\mathcal{G}}) \\ \Psi &\mapsto \tilde \Psi ,\end{align*} and since $$\Psi$$ was an isomorphism, so is $$\tilde \Psi$$.

It remains to handle the $$n\geq 2$$, case when (say) $$U = U_1 {\textstyle\coprod}U_2$$ has more than 1 connected component. Actually, is it even true that adjunctions preserve isomorphisms…? Todo: help??

Alternatively, consider the map $$\Psi$$ defined on presheaves – by the universal property, we get some sheaf morphism $$\tilde\Psi$$, which we can show is an isomorphism by showing its induced map on stalks is an isomorphism. This amounts to showing the following map is a group isomorphism: \begin{align*} \Psi_p: (\underline{A}^{\mathsf{pre}}({-}))_p \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathsf{Top}}({-}, A)_p .\end{align*}

First we identify the LHS: \begin{align*} (\underline{A}^{\mathsf{pre}}({-}))_p \coloneqq\colim_{U\ni p} \underline{A}^{\mathsf{pre}}(U) = \colim_{U\ni p} A = A .\end{align*}

#todo: show $$A$$ satisfies the universal property for a colimit.

Identifying the RHS, we have equivalence classes $$[U\ni p, s: U\to A]$$

• Injectivity: that $$\Psi_p$$ is injective follows from the fact that $$\ker \psi_p \coloneqq\left\{{a\in A {~\mathrel{\Big\vert}~}\Psi_p(a) = e}\right\}$$, where $$e$$ is the identity in the right-hand side stalk, which is represented by the class $$[U, f_e:U\to A]$$ where $$f_e(x) \coloneqq e_A$$, the identity of $$A$$, for every $$x\in U$$.

• Surjectivity: that $$\Psi_p$$ is surjective follows from the fact that every fixed $$f: U\to A$$ for $$A$$ discrete is constant on connected components. Use that $$p$$ is contained in a connected component $$U_1 \ni p$$, then $$[U, f] \sim [U_1, { \left.{{f}} \right|_{{U_1}} }] \coloneqq[U_1, g]$$ to get that $$g$$ is now a constant function of $$U_1$$. So $$g(x) = a$$ for some $$a\in A$$, so $$g = \Psi_p(a)$$ is in the image.

Alternatively:

• Show that $$\underline{A}$$ satisfies the universal property of $$(\underline{A}^{\mathsf{pre}})^+$$: we need to produce a morphism $$\theta: (\underline{A}^{\mathsf{pre}}) \to \underline{A}$$ such that for any $${\mathcal{G}}\in {\mathsf{Sh}}(X)$$ and morphism of presheaves $$\varphi: \underline{A}^{\mathsf{pre}}\to { \left.{{{\mathcal{G}}}} \right|_{{{\mathsf{pre}}}} }$$ we can produce a unique morphism $$\tilde \varphi$$ of sheaves making the following diagram commute:

• To define $$\tilde \varphi$$, it suffices to define morphisms of the form \begin{align*} \tilde\varphi(U): \underline{A}(U) &\to{\mathcal{G}}(U) \\ f & \mapsto \tilde\varphi(U)(f) \end{align*}

• Take a map $$f\in \underline{A}(U) \coloneqq\mathop{\mathrm{Hom}}_{\mathsf{Top}}(U, A)$$. Write $$U \coloneqq{\textstyle\coprod}U_i$$ as a union of connected components. Use that $$f$$ is constant on connected components since $$A$$ is discrete, so $$f(U_i) = a_i$$ for some elements $$a_i \in A \in {\mathsf{Ab}}{\mathsf{Grp}}$$.

• Plug the $$U_i$$ into $$\underline{A}^{\mathsf{pre}}$$ to get morphisms \begin{align*} \varphi(U_i): \underline{A}^{\mathsf{pre}}(U_i)= A \to { \left.{{{\mathcal{G}}}} \right|_{{{\mathsf{pre}}}} }(U_i) && \in {\mathsf{Ab}}{\mathsf{Grp}} \end{align*}

• Write $$b_i \coloneqq\varphi(U_i)(a_i) \in { \left.{{{\mathcal{G}}}} \right|_{{{\mathsf{pre}}}} }(U_i) = {\mathcal{G}}(U_i)$$.

• Since $${\mathcal{G}}$$ is in fact a sheaf, by unique gluing there exists a unique element $$b \in {\mathcal{G}}(U)$$ such that $${ \left.{{b}} \right|_{{U_i}} } = b_i$$. So define $$\tilde\varphi(U)(f) \coloneqq b$$.

• Now define the map $$\theta: \underline{A}^{\mathsf{pre}}(U_i) \to \mathop{\mathrm{Hom}}(U_i, A)$$ sending $$a_i$$ to the constant function $$f_{i}(x)\coloneqq a_i$$. Since $$\underline{A}$$ is a sheaf, there is a well defined $$F\in \mathop{\mathrm{Hom}}(U, A)$$ such that $${ \left.{{F}} \right|_{{U_i}} } = f_i$$. So for $$a\in \underline{A}^{\mathsf{pre}}(U)$$ set $$\theta(a) = F \in \underline{A}(U)$$.

• This makes the relevant diagram commute: if $$a\in A = \underline{A}^{\mathsf{pre}}(U)$$, then $$b\coloneqq\phi(U)(a) \in {\mathcal{G}}(U)$$. On the other hand, $$\theta(a)$$ is the constant function $$f_a: x\mapsto a$$ (on every connected component of $$U$$), and setting $$F \coloneqq\tilde\phi(f_a)\in {\mathcal{G}}(U)$$, we have $$F \coloneqq b$$.

• For any morphism of sheaves $$\varphi: {\mathcal{F}}\rightarrow {\mathcal{G}}$$, show that for each point $$p$$ that $$\ker (\varphi)_{p}=$$ $$\operatorname{ker}\left(\varphi_{p}\right)$$ and $$\operatorname{im}(\varphi)_{p} = \operatorname{im}\left(\varphi_{p}\right)$$.

• Show that $$\varphi$$ is injective (resp. surjective) if and only if the induced map on the stalks $$\varphi_{p}$$ is injective (resp. surjective) for all $$p$$.

• Show that a sequence of sheaves and morphisms \begin{align*} \cdots {\mathcal{F}}^{i-1} \xrightarrow{\varphi^{i-1}} {\mathcal{F}}^i \xrightarrow{\varphi^{i}} {\mathcal{F}}^{i+1} \to \cdots \end{align*} is exact if and only if for each $$P \in X$$ the corresponding sequence of stalks is exact as a sequence of abelian groups.
\envlist

• Write $$K\in {\mathsf{Sh}}(X)$$ for the kernel sheaf $$U \mapsto \ker \qty{ {\mathcal{F}}(U) \xrightarrow{\phi(U)} {\mathcal{G}}(U) }$$,
• We then want to show $$K_p = \ker\qty{{\mathcal{F}}_p \xrightarrow{\phi_p} {\mathcal{G}}_p}$$, an equality of sets in $${\mathsf{Ab}}{\mathsf{Grp}}$$. So we just do it!
• Addendum: this works because both are subsets of the same abelian group, $${\mathcal{F}}_p$$.
• We can write \begin{align*} \phi_p: {\mathcal{F}}_p &\to {\mathcal{G}}_p \\ [U, s] &\mapsto [U, \phi(U)(s)] ,\end{align*} and note that the zero element in a stalk is the equivalence class $$[U, 0]$$ where $$0\in {\mathsf{Ab}}{\mathsf{Grp}}$$ is the zero object. Thus \begin{align*} \ker \phi_p &\coloneqq\left\{{ x\in {\mathcal{F}}_p {~\mathrel{\Big\vert}~}\phi_p(x) = 0 \in {\mathcal{G}}_p }\right\} \\ & = \left\{{ [U, s] \in {\mathcal{F}}_p {~\mathrel{\Big\vert}~}[U, \phi(U)(s)] = [U, 0] }\right\} \\ & = \left\{{ [U, s] \in {\mathcal{F}}_p {~\mathrel{\Big\vert}~}\phi(U)(s) = 0 }\right\} \\ & = \left\{{ [U, s] \in {\mathcal{F}}_p {~\mathrel{\Big\vert}~}s \in \ker \phi(U) }\right\} \\ &= \left\{{ [U, s] {~\mathrel{\Big\vert}~}s\in \ker{\qty{{\mathcal{F}}(U) \xrightarrow{\phi(U)} {\mathcal{G}}(U) } } }\right\} \\ &\coloneqq\left\{{[U, s] {~\mathrel{\Big\vert}~}s\in K(U)}\right\} \\ &\coloneqq K_p .\end{align*}
\envlist

• Write $${\mathcal{I}}$$ for the sheaf $$\operatorname{im}\phi$$ which sends $$U\mapsto \operatorname{im}\qty{{\mathcal{F}}(U) \xrightarrow{\phi(U)} {\mathcal{G}}(U)}$$.
• We want to show $${\mathcal{I}}_p = \operatorname{im}\qty{{\mathcal{F}}_p \xrightarrow{\phi_p} {\mathcal{G}}_p}$$, where both are subsets of $${\mathcal{G}}_p$$.
• So we show set equality: \begin{align*} \operatorname{im}(\phi_p) &= \left\{{ y\in {\mathcal{G}}_p {~\mathrel{\Big\vert}~}\exists x\in {\mathcal{F}}_p,\, \phi_p(x) = y }\right\} \\ &= \left\{{ [U, t] \in {\mathcal{G}}_p {~\mathrel{\Big\vert}~}\exists [U, s] \in {\mathcal{F}}_p,\, \phi_p([U, s]) = [U, t] }\right\} \\ &= \left\{{ [U, t] \in {\mathcal{G}}_p {~\mathrel{\Big\vert}~}\exists s\in {\mathcal{F}}(U),\, \phi(U)(s) = t }\right\} \\ &= \left\{{ [U, t] \in {\mathcal{G}}_p {~\mathrel{\Big\vert}~}t\in \operatorname{im}\qty{{\mathcal{F}}(U) \xrightarrow{\phi(U)} {\mathcal{G}}(U) }}\right\} \\ &\coloneqq\left\{{ [U, t] \in {\mathcal{G}}_p {~\mathrel{\Big\vert}~}t \in {\mathcal{I}}(U) }\right\} \\ &\coloneqq{\mathcal{I}}_p .\end{align*}

$$\implies$$:

• Use that injectivity of a morphism $$\phi$$ of sheaves is defined to hold exactly when $$\ker \phi = 0$$ is the constant zero sheaf.

• Now use (1): \begin{align*} 0 = \ker(\phi) \implies 0 = \ker(\phi)_p = \ker(\phi_p) && \forall p .\end{align*}

• If $$\ker \phi = 0$$, so $$\phi$$ is injective, then $$\ker \phi_p = 0$$ for all $$p$$, so $$\ker \phi_p$$ is injective.

$$\impliedby$$:

• Conversely, suppose $$\ker \phi_p = 0$$ for all $$p$$; we want to show $$\ker \phi(U) = 0$$ for all $$U$$.
• So fix $$U\ni p$$, we want to show \begin{align*} s\in K(U) \coloneqq\ker\qty{{\mathcal{F}}(U) \xrightarrow{\phi(U)} {\mathcal{G}}(U)} \implies s = 0 \in {\mathcal{F}}(U) .\end{align*}
• We have $$\phi(U)(s) = 0$$, so \begin{align*} \phi_p([U, s]) \coloneqq[U, \phi(U)(s)] = [U, 0] \in {\mathcal{G}}_p \implies [U, s] \in \ker (\phi_p) .\end{align*}
• By injectivity of $$\phi_p$$, we have $$[U, s] = 0 \in {\mathcal{F}}_p$$.
• So there is some open $$U_p$$ with $$U \supseteq U_p \ni p$$ and $$\mathop{\mathrm{Res}}(U, U_p)(s) = 0$$ in $${\mathcal{F}}(U_p)$$.
• Then $$\left\{{U_p }\right\}_{p\in U} \rightrightarrows U$$, and since $${\mathcal{F}}$$ is a sheaf, by existence of gluing these glue to an $$F \in {\mathcal{F}}(U)$$ with $$\mathop{\mathrm{Res}}(U, U_p)(F) = 0$$ for each $$p$$. By uniqueness of gluing, $$0 = F = s$$.

$$\implies$$:

• Suppose $$\phi$$ is surjective, then by definition $$\operatorname{im}\phi = {\mathcal{G}}$$ is an equality of sheaves.
• So $$(\operatorname{im}\phi)(U) = {\mathcal{G}}(U)$$ for all $$U$$.
• Let $$[U, t]\in {\mathcal{G}}_p$$, so $$t\in {\mathcal{G}}(U)$$.
• Then $$t\in (\operatorname{im}\phi)(U)$$, so there exists an $$s\in {\mathcal{F}}(U)$$ such that $$\phi(U)(s) = t$$.
• Then $$[U, s] \mapsto [U, \phi(U)(s)] = [U, t]$$, under $$\phi_p$$, making $$\phi_p$$ surjective.

$$\impliedby$$:

• Suppose $$\phi_p$$ is surjective for all $$p$$, fix $$U$$, and let $$t \in {\mathcal{G}}(U)$$. We want to produce an $$s\in {\mathcal{F}}(U)$$ such that $$\phi(U)(s) = t$$.

• For $$p\in U$$, the image of $$t$$ in the stalk of $${\mathcal{G}}$$ is of the form $$[U_p, t_p] \in {\mathcal{G}}_p$$ where $$t_p \in {\mathcal{G}}(U_p)$$.

• Since $$\phi_p: {\mathcal{F}}_p \twoheadrightarrow{\mathcal{G}}_p$$, we can find some pair $$[U_p, s_p]$$ mapping to $$[U_p, t_p]$$ under $$\phi_p$$, so $$\phi(U_p)(s_p) = t_p$$.

• Note that $$\mathop{\mathrm{Res}}(U, U_p)(t) = t_p$$.
• Note: may need to pull back to some $$\tilde U_p$$, then take a common refinement in both germs?
• Now $$\left\{{U_p}\right\}_{p\in U}\rightrightarrows U$$, so using existence of gluing for $${\mathcal{F}}$$ we have some $$s\in {\mathcal{F}}(U)$$ with $$\mathop{\mathrm{Res}}(U, U_p)(s) = s_p$$ for all $$p$$.

• Claim: $$\phi(U)(s) = t$$. \begin{align*} \mathop{\mathrm{Res}}(U, U_p)( \phi(s) ) &= \phi(\mathop{\mathrm{Res}}(U, U_p)(s)) \\ &= \phi(s_p) \\ &= t_p \\ &= \mathop{\mathrm{Res}}(U, U_p)(t) && \forall p\in U ,\end{align*} so $$\phi(s) = t$$ by uniqueness of gluing of $${\mathcal{G}}$$.

$$\implies$$: Assuming exactness of sheaves, \begin{align*} \ker({\mathcal{F}}^{i+1}) = \operatorname{im}({\mathcal{F}}^{i}) \iff \ker({\mathcal{F}}^{i+1})_p = \operatorname{im}({\mathcal{F}}^{i})_p && \forall p .\end{align*}

$$\impliedby$$: Assuming exactness on stalks, write \begin{align*} \ker({\mathcal{F}}^{i+1})_p &= \ker({\mathcal{F}}^{i+1}_p) && \text{by 1 } \\ &= \operatorname{im}({\mathcal{F}}^{i}_p) && \text{exactness, by assumption} \\ &= \operatorname{im}({\mathcal{F}}^{i})_p && \text{by 1} .\end{align*}

• Let $$\varphi: {\mathcal{F}}\to{\mathcal{G}}$$ be a morphism of sheaves on $$X$$. Show that $$\varphi$$ is surjective if and only if the following condition holds:

For every open set $$U \subseteq X$$, and for every $$s\in {\mathcal{G}}(U)$$, there is a cover $$\left\{{U_i}\right\}$$ of $$U$$ and elements $$t_i \in {\mathcal{F}}(U_i)$$ such that $$\varphi(t_i) = { \left.{{s}} \right|_{{U_i}} }$$ for all $$i$$.

• Give an example of a surjective morphism of sheaves $$\varphi: {\mathcal{F}}\rightarrow {\mathcal{G}}$$, and an open set $$U$$ such that $$\varphi(U): {\mathcal{F}}(U) \rightarrow {\mathcal{G}}(U)$$ is not surjective.

$$\implies$$:

• If $$\phi: {\mathcal{F}}\twoheadrightarrow{\mathcal{G}}$$, then $$\phi_p: {\mathcal{F}}_p \twoheadrightarrow{\mathcal{G}}_p$$ for all $$p$$, since $$\operatorname{im}(\phi_p) = (\operatorname{im}\phi)_p = {\mathcal{G}}_p$$, using problem 1.2.
• Fix $$U \subseteq X$$ and $$s\in {\mathcal{G}}(U)$$, we want
• To produce a cover $$\left\{{U_i}\right\}\rightrightarrows U$$,
• To find $$t_i\in {\mathcal{F}}(U_i)$$, and
• To show that $$\phi(t_i) = \mathop{\mathrm{Res}}(U, U_i)(s)$$ for all $$i$$.
• Fix $$p$$, and take the image of $$s$$ in the stalk of $${\mathcal{G}}$$ to get $$[U_p, s_p] \in {\mathcal{G}}_p$$ with $$s_p \in {\mathcal{G}}(U_p)$$ and $$\mathop{\mathrm{Res}}(U, U_p)(s) = s_p$$. Note that $$\left\{{U_p}\right\}_{p\in U}\rightrightarrows U$$.
• By surjectivity on stalks, these pull back to $$[U_p, t_p]\in {\mathcal{F}}_p$$ with $$t_p \in {\mathcal{F}}(U_p)$$ and $$\phi_p([U_p, t_p]) \coloneqq[U_p, \phi(U_p)(t_p)] = [U_p, s_p]$$.
• Then $$s_p \in \operatorname{im}({\mathcal{F}}(U_p) \xrightarrow{\phi(U_p)} {\mathcal{G}}(U_p ))$$ and $$\phi(t_p) = s_p = \mathop{\mathrm{Res}}(U, U_p)(s)$$.

$$\impliedby$$:

• If $$\left\{{U_i}\right\}\rightrightarrows U$$ with $$\phi(t_i) = \mathop{\mathrm{Res}}(U, U_i)(s)$$ for all $$i$$, then the $$t_i$$ glue to a unique section $$t\in {\mathcal{F}}(U)$$ since $${\mathcal{F}}$$ is a sheaf.
• Moreover $$\mathop{\mathrm{Res}}(U, U_i)( \phi(t) ) = \phi(\mathop{\mathrm{Res}}(U, U_i)(t)) = \phi(t_i) = \mathop{\mathrm{Res}}(U, U_i)(s)$$ for all $$i$$, and by unique gluing for $${\mathcal{G}}$$ we have $$\phi(t) = s$$.
• So $$\phi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U)$$ is surjective for all $$U$$, making $$\operatorname{im}(\phi(U)) = {\mathcal{G}}(U)$$
• So $$\operatorname{im}\phi = {\mathcal{G}}$$ as sheaves since they make the same assignment to every open set $$U$$, making $$\phi: {\mathcal{F}}\to{\mathcal{G}}$$ surjective by definition.
\envlist

• Take $$X \coloneqq\left\{{a,b,c}\right\}$$ a 3-point space with the topology $$\tau_X \coloneqq\left\{{\emptyset, \left\{{a}\right\}, \left\{{b}\right\}, \left\{{a,b}\right\}, X}\right\}$$.

• Take $${\mathcal{F}}\coloneqq\underline{A}$$ for some nontrivial $$A\in {\mathsf{Ab}}{\mathsf{Grp}}$$. We have the stalks

• $${\mathcal{F}}_a = A$$
• $${\mathcal{F}}_b = A$$
• $${\mathcal{F}}_c = A$$
• Take $${\mathcal{G}}\coloneqq\underline{A}(a) \times \underline{A}(b)$$, the skyscraper sheaves at $$a$$ and $$b$$ respectively, where \begin{align*} \underline{A}(x)(U) \coloneqq \begin{cases} A & x\in U \\ 0 & \text{ else} . \end{cases} \end{align*} Note that the stalks are given by $$\underline{A}(x)_x = A$$ and $$\underline{A}(x)_y = 0$$ for $$y\neq x$$, so

• $${\mathcal{G}}_a = A\times 0$$
• $${\mathcal{G}}_b = 0 \times A$$
• $${\mathcal{G}}_c = 0 \times 0$$.
• Now define $$\phi: {\mathcal{F}}\to {\mathcal{G}}$$ by specifying $$\phi(U):{\mathcal{F}}(U) \to {\mathcal{G}}(U)$$ for all $$U$$ in the following way:

• Note that the induced maps on stalks are surjective, since $$\phi_p: A \to A, 0$$ is either the identity or the zero map. But e.g. for $$\left\{{a, b}\right\}$$ we have $$A\mapsto A^{\times 2}$$, which can not be surjective.

Question: what is this map? Apparently its image is the diagonal…?

• Let $$\varphi: {\mathcal{F}}\to {\mathcal{G}}$$ be a morphism of presheaves such that $$\varphi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U)$$ is injective for each $$U$$. Show that the induced map $$\varphi^+: {\mathcal{F}}^+ \to {\mathcal{G}}^+$$of associated sheaves is injective.

• Use part (a) to show that if $$\varphi: {\mathcal{F}}\to{\mathcal{G}}$$ is a morphism of sheaves, then $$\operatorname{im}\varphi$$ can be naturally identified with a subsheaf of $${\mathcal{G}}$$, as mentioned in the text.

\envlist

• $$\phi: {\mathcal{F}}\to {\mathcal{G}}$$ is injective iff $$\phi_p:{\mathcal{F}}_p \to {\mathcal{G}}_p$$ is injective for all $$p$$.
• Sheafification induces a map $$\phi^+: {\mathcal{F}}^+_p \to {\mathcal{G}}^+_p$$
• The sheafification has the same stalks, so $${\mathcal{F}}^+_p = {\mathcal{F}}_p$$ and $${\mathcal{G}}^+_p = {\mathcal{G}}_p$$.
• So in fact $$\phi^+_p = \phi_p$$. Since $$\phi^+_p$$ is thus injective on all stalks, $$\phi^+$$ is injective on sheaves.
\envlist

• Noting that on opens $$(\operatorname{im}\phi)(U) \subseteq {\mathcal{G}}(U)$$ is an inclusion of abelian groups, so define a morphism of sheaves by $$\iota(U): (\operatorname{im}\phi)(U) \to {\mathcal{G}}(U)$$ using this inclusion.
• By definition, it suffices to show $$\ker \iota = 0$$ as a sheaf.
• By 1.2.2, it suffices to show $$(\ker \iota)_p = 0$$ on all stalks.
• By 1.2.1, $$(\ker \iota)_p = \ker(\iota_p)$$, so it suffices to show $$\iota_p$$ is injective for all $$p$$.
• Now use that \begin{align*} \ker(\iota_p) = \colim_{U\ni p} (\ker \phi)(\iota(U)) = \colim_{U\ni p} 0 = 0 ,\end{align*} since all of the $$\iota(U)$$ are injective, so 0 satisfies the universal property for this colimit. So we’re done.

Show that a morphism of sheaves is an isomorphism if and only if it is both injective and surjective.

\envlist


Problem: surjections of sheaves don’t induce surjections ons sections!

• $$\phi:{\mathcal{F}}\to{\mathcal{G}}$$ being injective means that $$(\ker \phi) = 0$$ as sheaves, and surjective means $$(\operatorname{im}\phi) = {\mathcal{G}}$$.

• Thus $$\phi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U)$$ is injective, since $$(\ker \phi)(U) = 0(U) = 0$$, and surjective since $$\operatorname{im}(\phi(U)) = (\operatorname{im}\phi)(U) = {\mathcal{G}}(U)$$. This $$\phi(U)$$ is an isomorphism in abelian groups, and has an left and right inverse $$\phi^{-1}(U): {\mathcal{G}}(U) \to {\mathcal{F}}(U)$$.

• So we have a diagram:

• Both squares form a morphism of sheaves, so the right square assembles to $$\phi^{-1}: {\mathcal{G}}\to{\mathcal{F}}$$
• Moreover $$(\phi^{-1}\circ \phi)({\mathcal{F}})(U) = \operatorname{id}_{{\mathcal{F}}(U)}$$ and similarly in the other order, so $$\phi^{-1}\circ \phi= \operatorname{id}_{{\mathcal{F}}}$$ Similarly $$(\phi\circ \phi^{-1})({\mathcal{G}})(U) = \operatorname{id}_{{\mathcal{G}}(U)}$$ and $$(\phi^{-1}\circ \phi) = \operatorname{id}_{{\mathcal{G}}}$$.
• Then by definition an isomorphism of sheaves is a morphism with a two-sided inverse, so we’re done.
#todo