In class, we defined a morphism $\phi: \mathcal{F} \rightarrow \mathcal{G}$ of presheaves to be injective if the map $\phi(U): \mathcal{F}(U) \rightarrow \mathcal{G}(U)$ is injective for every open $U $ in the topological space $X$.
Now there is an exercise, which goes as:
Problem: Let $\phi: \mathcal{F} \rightarrow \mathcal{G}$ be a morphism of presheaves of sets on the topological space $X$. Show that $\phi$ is a monomorphism in the category of presheaves of sets on $X$ iff $\phi$ is injective as a morphism of presheaves.
I am a little bit confused of what this means. I know what a monomorphism means in a category $\mathcal{C}$. We call an arrow $f: A \rightarrow B$ between objects in the category $\mathcal{C}$ a monomorphism, if for any two arrows $g_1, g_2: C \rightarrow A$ with $f \circ g_1 = f \circ g_2$ it follows that $g_1 = g_2$.
My confusion is: what does the statement "$\phi$ is a monomorphism in the category of presheaves of sets on $X$" mean? Are the presheaves $\mathcal{F}$ and $\mathcal{G}$ considered as the objects here? So does $\phi$ play the role of $f$ in the definition of monomorphism given earlier?
If that is the case, I have no idea how to prove this. I assume we have a monomorphism. Let $U \subset X$ be open. I wish to show that $\phi(U) : \mathcal{F}(U) \rightarrow \mathcal{G} (U)$ is injective. So given sections $s_1, s_2 \in \mathcal{F}(U)$ with $\phi(U)(s_1) = \phi(U)(s_2)$, I wish to prove that $s_1 = s_2$. How can this be done?