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
\[ \xymatrix{ \mathcal{F} \ar@<-1ex>[r] \ar@<1ex>[r] & \mathcal{F}' \\ f^{-1}f_*\mathcal{F} \ar@<-1ex>[r] \ar@<1ex>[r] \ar[u] & f^{-1}f_*\mathcal{F}' \ar[u] } \]
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
\[ \xymatrix{ \mathcal{F} \ar[r] & \mathcal{F}' \\ f^{-1}f_*\mathcal{F} \ar[r] \ar[u] & f^{-1}f_*\mathcal{F}' \ar[u] } \]
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
\[ f_*\mathcal{F}' \amalg _{f_*\mathcal{F}} f_*\mathcal{F}' = f_*(\mathcal{F}' \amalg _\mathcal {F} \mathcal{F}') \longrightarrow f_*\mathcal{F}' \]
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
\[ f_*(\mathcal{F} \times _{\mathcal{F}'} \mathcal{F}) = f_*\mathcal{F} \times _{f_*\mathcal{F}'} f_*\mathcal{F} \longrightarrow f_*\mathcal{F} \]
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
\[ \xymatrix{ f_*\mathcal{F} \ar[r] & f_*f^{-1}\mathcal{G} \\ \mathcal{G}' \ar[u] \ar[r] & \mathcal{G} \ar[u] } \]
so that $\mathcal{G}' \to \mathcal{G}$ is surjective also. Consider the commutative diagram
\[ \xymatrix{ \mathcal{F} \ar[r] & f^{-1}\mathcal{G} \\ f^{-1}f_*\mathcal{F} \ar[r] \ar[u] & f^{-1}f_*f^{-1}\mathcal{G} \ar[u] \\ f^{-1}\mathcal{G}' \ar[u] \ar[r] & f^{-1}\mathcal{G} \ar[u] } \]
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
\[ \xymatrix{ \mathcal{F} \ar[r] & \mathcal{F}' \\ \mathcal{F}'' \ar[u] \ar[r] & f^{-1}f_*\mathcal{F}' \ar[u] } \]
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 $\mathcal{F}'' \to f^{-1}f_*\mathcal{F}'$:
\[ \xymatrix{ & \mathcal{F} \ar[r] & \mathcal{F}' \\ f^{-1}\mathcal{G}' \ar[r] & \mathcal{F}'' \ar[u] \ar[r] & f^{-1}f_*\mathcal{F}' \ar[u] } \]
Pushing this down using $f_*$ we get a commutative diagram
\[ \xymatrix{ & f_*\mathcal{F} \ar[r] & f_*\mathcal{F}' \\ f_*f^{-1}\mathcal{G}' \ar[r] & f_*\mathcal{F}'' \ar[u] \ar[r] & f_*f^{-1}f_*\mathcal{F}' \ar[u] \\ \mathcal{G}' \ar[u] \ar[rr] & & f_*\mathcal{F}' \ar[u] } \]
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.
$\square$
Comments (0)