The Stacks project

Proof. We shall show that $\varphi : \mathcal{F} \to \mathcal{G}$ is injective if and only if it is a monomorphism of $\textit{PSh}(\mathcal{C})$. Indeed, the “only if” direction is straightforward, so let us show the “if” direction. Assume that $\varphi$ is a monomorphism. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$; we need to show that $\varphi _ U$ is injective. So let $a, b \in \mathcal{F}(U)$ be such that $\varphi _ U (a) = \varphi _ U (b)$; we need to check that $a = b$. Under the isomorphism (7.2.1.1), the elements $a$ and $b$ of $\mathcal{F}(U)$ correspond to two natural transformations $a', b' \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{F})$. Similarly, under the analogous isomorphism $\mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{G}) = \mathcal{G}(U)$, the two equal elements $\varphi _ U (a)$ and $\varphi _ U (b)$ of $\mathcal{G}(U)$ correspond to the two natural transformations $\varphi \circ a', \varphi \circ b' \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{G})$, which therefore must also be equal. So $\varphi \circ a' = \varphi \circ b'$, and thus $a' = b'$ (since $\varphi$ is monic), whence $a = b$. This finishes (1).

We shall show that $\varphi : \mathcal{F} \to \mathcal{G}$ is surjective if and only if it is an epimorphism of $\textit{PSh}(\mathcal{C})$. Indeed, the “only if” direction is straightforward, so let us show the “if” direction. Assume that $\varphi$ is an epimorphism.

For any two morphisms $f : A \to B$ and $g : A \to C$ in the category $\textit{Sets}$, we let $\text{inl}_{f,g}$ and $\text{inr}_{f,g}$ denote the two canonical maps from $B$ and $C$ to $B \coprod _ A C$. (Here, the pushout is evaluated in $\textit{Sets}$.)

Now, we define a presheaf $\mathcal{H}$ of sets on $\mathcal{C}$ by setting $\mathcal{H}(U) = \mathcal{G}(U) \coprod _{\mathcal{F}(U)} \mathcal{G}(U)$ (where the pushout is evaluated in $\textit{Sets}$ and induced by the map $\varphi _ U : \mathcal{F}(U) \to \mathcal{G}(U)$) for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$; its action on morphisms is defined in the obvious way (by the functoriality of pushout). Then, there are two natural transformations $i_1 : \mathcal{G} \to \mathcal{H}$ and $i_2 : \mathcal{G} \to \mathcal{H}$ whose components at an object $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ are given by the maps $\text{inl}_{\varphi _ U, \varphi _ U}$ and $\text{inr}_{\varphi _ U, \varphi _ U}$, respectively. The definition of a pushout shows that $i_1 \circ \varphi = i_2 \circ \varphi$, whence $i_1 = i_2$ (since $\varphi$ is an epimorphism). Thus, for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, we have $\text{inl}_{\varphi _ U, \varphi _ U} = \text{inr}_{\varphi _ U, \varphi _ U}$. Thus, $\varphi _ U$ must be surjective (since a simple combinatorial argument shows that if $f : A \to B$ is a morphism in $\textit{Sets}$, then $\text{inl}_{f,f} = \text{inr}_{f,f}$ if and only if $f$ is surjective). In other words, $\varphi$ is surjective, and (2) is proven.

We shall show that $\varphi : \mathcal{F} \to \mathcal{G}$ is both injective and surjective if and only if it is an isomorphism of $\textit{PSh}(\mathcal{C})$. This time, the “if” direction is straightforward. To prove the “only if” direction, it suffices to observe that if $\varphi$ is both injective and surjective, then $\varphi _ U$ is an invertible map for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and the inverses of these maps for all $U$ can be combined to a natural transformation $\mathcal{G} \to \mathcal{F}$ which is an inverse to $\varphi$. $\square$

Proof. To prove existence, just set $\mathcal{G}'(U) = \varphi _ U \left(\mathcal{F}(U)\right)$ for every $U \in \mathop{\mathrm{Ob}}\nolimits (C)$ (and inherit the action on morphisms from $\mathcal{G}$), and prove that this defines a subpresheaf of $\mathcal{G}$ and that $\varphi$ factors as $\mathcal{F} \to \mathcal{G}' \to \mathcal{G}$ with the first map being surjective. Uniqueness is straightforward. $\square$