The Stacks project

Proof. Proof of (a): This is immediate from the definitions.

Proof of (b). Suppose that $a, b : \mathcal{F} \to \mathcal{F}'$ are maps of sheaves on $\mathcal{C}$. If $f_*a = f_*b$, then $f^{-1}f_*a = f^{-1}f_*b$. Consider the commutative diagram

If the bottom two arrows are equal and the vertical arrows are surjective then the top two arrows are equal. Hence (b) follows.

Proof of (c). Suppose that $a : \mathcal{F} \to \mathcal{F}'$ is a map of sheaves on $\mathcal{C}$. Consider the commutative diagram

If (7) holds, then the vertical arrows are isomorphisms. Hence if $f_*a$ is injective (resp. surjective, resp. bijective) then the bottom arrow is injective (resp. surjective, resp. bijective) and hence the top arrow is injective (resp. surjective, resp. bijective). Thus we see that (7) implies (8), (9), (10). It is clear that (7) implies (3). The implications (7) $\Rightarrow$ (2), (1) follow from (a) and (h) which we will see below.

Proof of (d). Assume (3). Suppose that $a : \mathcal{F} \to \mathcal{F}'$ is a map of sheaves on $\mathcal{C}$ such that $f_*a$ is surjective. As $f^{-1}$ is exact this implies that $f^{-1}f_*a : f^{-1}f_*\mathcal{F} \to f^{-1}f_*\mathcal{F}'$ is surjective. Combined with (3) this implies that $a$ is surjective. This means that (9) holds. Assume (9). Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$. We have to show that the map $f^{-1}f_*\mathcal{F} \to \mathcal{F}$ is surjective. It suffices to show that $f_*f^{-1}f_*\mathcal{F} \to f_*\mathcal{F}$ is surjective. And this is true because there is a canonical map $f_*\mathcal{F} \to f_*f^{-1}f_*\mathcal{F}$ which is a one-sided inverse.

Proof of (e). We use Categories, Lemma 4.13.3 without further mention. If $\mathcal{F} \to \mathcal{F}'$ is surjective then $\mathcal{F}' \amalg _\mathcal {F} \mathcal{F}' \to \mathcal{F}'$ is an isomorphism. Hence (6) implies that

is an isomorphism also. And this in turn implies that $f_*\mathcal{F} \to f_*\mathcal{F}'$ is surjective. Hence we see that (6) implies (4). If $\mathcal{F} \to \mathcal{F}'$ is surjective then $\mathcal{F}'$ is the coequalizer of the two projections $\mathcal{F} \times _{\mathcal{F}'} \mathcal{F} \to \mathcal{F}$ by Lemma 7.11.3. Hence if (5) holds, then $f_*\mathcal{F}'$ is the coequalizer of the two projections

which clearly means that $f_*\mathcal{F} \to f_*\mathcal{F}'$ is surjective. Hence (5) implies (4) as well.

Proof of (f). Assume (4). Let $\mathcal{F} \to f^{-1}\mathcal{G}$ be a surjective map of sheaves on $\mathcal{C}$. By (4) we see that $f_*\mathcal{F} \to f_*f^{-1}\mathcal{G}$ is surjective. Let $\mathcal{G}'$ be the fibre product

so that $\mathcal{G}' \to \mathcal{G}$ is surjective also. Consider the commutative diagram

and we see the required result. Conversely, assume (11). Let $a : \mathcal{F} \to \mathcal{F}'$ be surjective map of sheaves on $\mathcal{C}$. Consider the fibre product diagram

Because the lower horizontal arrow is surjective and by (11) we can find a surjection $\gamma : \mathcal{G}' \to f_*\mathcal{F}'$ such that $f^{-1}\gamma$ factors through $\mathcal{F}'' \to f^{-1}f_*\mathcal{F}'$:

Pushing this down using $f_*$ we get a commutative diagram

which proves that (4) holds.

Proof of (g). Assume (9). We use Categories, Lemma 4.13.3 without further mention. Let $a : \mathcal{F} \to \mathcal{F}'$ be a map of sheaves on $\mathcal{C}$ such that $f_*a$ is injective. This means that $f_*\mathcal{F} \to f_*\mathcal{F} \times _{f_*\mathcal{F}'} f_*\mathcal{F} = f_*(\mathcal{F} \times _{\mathcal{F}'} \mathcal{F})$ is an isomorphism. Thus by (9) we see that $\mathcal{F} \to \mathcal{F} \times _{\mathcal{F}'} \mathcal{F}$ is surjective, i.e., an isomorphism. Thus $a$ is injective, i.e., (8) holds. Since (10) is trivially equivalent to (8) $+$ (9) we are done with (g).

Proof of (h). This is Categories, Lemma 4.24.4. $\square$

Proof. Let $a : \mathcal{F} \to \mathcal{G}$ be a surjective map of sheaves on $\mathcal{D}$. Let $U$ be an object of $\mathcal{C}$ and let $s \in f_*\mathcal{G}(U) = \mathcal{G}(u(U))$. By assumption there exists a covering $\{ V_ j \to u(U)\}$ and sections $s_ j \in \mathcal{F}(V_ j)$ with $a(s_ j) = s|_{V_ j}$. Now we may think of the sections $s$, $s_ j$ and $a$ as giving a commutative diagram of maps of sheaves

By assumption there exists a covering $\{ U_ i \to U\}$ such that we can enlarge the commutative diagram above as follows

Because $\mathcal{F}$ is a sheaf the map from the left lower corner to the right upper corner corresponds to a family of sections $s_ i \in \mathcal{F}(u(U_ i))$, i.e., sections $s_ i \in f_*\mathcal{F}(U_ i)$. The commutativity of the diagram implies that $a(s_ i)$ is equal to the restriction of $s$ to $U_ i$. In other words we have shown that $f_*a$ is a surjective map of sheaves. $\square$

Proof. Let $\alpha : \mathcal{F} \to \mathcal{G}$ be maps of sheaves on $\mathcal{D}$. By assumption for every object $V$ of $\mathcal{D}$ we get $\mathcal{F}(V) \subset \prod \mathcal{F}(u(U_ i)) = \prod f_*\mathcal{F}(U_ i)$ by the sheaf condition for some $U_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and similarly for $\mathcal{G}$. Hence it is clear that if $f_*\alpha$ is injective, then $\alpha$ is injective. In other words $f_*$ reflects injections.

Suppose that $f_*\alpha$ is surjective. Then for $V, U_ i, u(U_ i) \to V$ as above and a section $s \in \mathcal{G}(V)$, there exist coverings $\{ U_{ij} \to U_ i\}$ such that $s|_{u(U_{ij})}$ is in the image of $\mathcal{F}(u(U_{ij}))$. Since $\{ u(U_{ij}) \to V\}$ is a covering (as $u$ is continuous and by the axioms of a site) we conclude that $s$ is locally in the image. Thus $\alpha$ is surjective. In other words $f_*$ reflects surjections. $\square$