Exact sequence of groups to exact sequence sheaves

Question

For a topological group $G$ and a topological space $X$, denote by $\underline{G^X}$ the sheaf of continuous functions from $X$ into $G$.

Suppose we have an exact sequence of groups $$ 1\rightarrow F\rightarrow G\rightarrow H\rightarrow 1. $$ What are sufficient conditions for the corresponding sequence of sheaves $$ 1\rightarrow \underline{F^X}\rightarrow \underline{G^X}\rightarrow \underline{H^X}\rightarrow 1 $$ to be exact?

Answer 1

Exactness at $\underline{F^X}$ and $\underline{G^X}$ is clear, so you are asking about surjectivity at $\underline{H^X}$. In other words, given a point $x \in X$, an open set $U \ni x$ and a continuous map $\phi : U \to H$, when can I shrink $U$ to $V \ni x$ and lift $\phi$ to $\psi: V \to G$?

I think the right condition is

There is a nonempty open set $W$ in $H$, and a continuous section $\sigma: W \to G$ of the map $G \to H$.

This implies that $\underline{G^X} \to \underline{H^X}$ is surjective. Proof: Let $(x, U, \phi)$ be as above. Using the group structure on $G$ and $H$, we may translate $W$ to contain $\phi(x)$ and adjust our section to still be a section. Now, take $V = \phi^{-1}(W)$. Since $\phi$ is continuous, $V$ is open, and $\sigma \circ \phi$ gives the desired lift of $\phi$. $\square$

In the other direction, if the boxed condition does not hold, then the identity map in $\underline{H^H}$ is not in the image of $\underline{G^H}$.

EDIT Oh, I see that Eric Wofsey has written the same thing back on MO.