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:

Link to 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 \(V\):

Link to Diagram

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.


  • 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:

Link to Diagram

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

Link to Diagram

  • 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.
  • 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*}
  • 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*}


  • 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.


  • 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, \(0 = F = s\).


  • 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.


  • 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.


  • 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 \(\phi(t_p) = s_p = \mathop{\mathrm{Res}}(U, U_p)(s)\).


  • 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.
  • 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:

Link to Diagram

  • 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.

  • \(\phi: {\mathcal{F}}\to {\mathcal{G}}\) 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.
  • 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.


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:

Link to 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.