Let $\mathcal F$ and $\mathcal G$ be sheaves of sets on a topological space $X$ and let $m: \mathcal F\to \mathcal G$ be a monomorphism, which simply means that $m_U: \mathcal F(U)\to \mathcal G(U)$ is injective for all opens $U\subseteq X$. Does $m$ always have a retraction?
I came up with the following idea: Fix a global section $a_0\in \mathcal F(X)$ and define $n: \mathcal G\to \mathcal F$ given by $$n_U(x) = \left\{\begin{array}{ll}a & m_U(a) = x \\ a_0|_U & x\not\in m_U(\mathcal{F}(U)).\end{array}\right.$$ If this $n$ is a morphism of sheaves, then it the wanted retraction. We need to show that $$ \begin{array}{cc} \mathcal G(U)\xrightarrow{n_U} & \mathcal F(U)\\ \downarrow & \downarrow \\ \mathcal G(V)\xrightarrow{n_V} & \mathcal F(V) \end{array}$$ commutes for all opens $V\subseteq U$. Now suppose that $x\in \mathcal G(U)$ and $x\not\in m_U(\mathcal F(U))$, then $n_U(x)|_V = a_0|_V$, so for the diagram to commute we must have $n_V(x|_V) = a_0|_V$. I haven't been able to prove this. Is this true at all? I think it comes down to whether $x\not \in m_U(\mathcal F(U))\implies x|_V \not\in m_V(\mathcal F(V))$ is true or not.
This is not true in general. For example, suppose $X = \mathbb{R}$ with $\mathcal{F} = (0, 1)$ and $\mathcal{G} = \mathbb{R}$. (In other words, $\mathcal{F}(U) = \{ 0 \}$ if $U \subseteq (0, 1)$, and $\mathcal{F}(U) = \emptyset$ otherwise; and $\mathcal{G}(U) = \{ 0 \}$ for all $U$.) Then there is a unique injection $\mathcal{F} \to \mathcal{G}$; however, there is no morphism $\mathcal{G} \to \mathcal{F}$ at all, since $\mathcal{F}(\mathbb{R}) = \emptyset$ has nowhere to send the global section $0 \in \mathcal{G}(\mathbb{R})$.