Proving this function is an isomorphism

Question

Let $1\longrightarrow H\longrightarrow G\longrightarrow F\longrightarrow 1$ be a short exact sequence with $H$ abelian. Let $T=\{t_{\sigma}\}$ be a transversal of $H$ in $G$. Let also be $Z^{1}(F,H)$ the 1-cocycles defined as normalized cochains. Given $d\in Z^{1}(F,H)$ I want to show that the map: $$\varphi(t_{\sigma}a):=t_{\sigma}d(\sigma)a$$ with $a\in H$ defines an isomorphism of $G$. If $F$ acts by conjugation then I know already how to prove it, but that'a an assumption which is nowehere else required.

Answer 1

1-Cocycles are always defined with respect to some action, in this way the notation can be somewhat misleading. $Z^1(F,H)$ is really $Z^1(F,H,\tau)$ where $\tau : F \to \text{Aut}(H)$ is an action.

This is because the cocycle condition, which says

$$d(g_1 g_2) = d(g_1) + g_1 \cdot d(g_2) = d(g_1) + \tau(g_1)(d(g_2))$$

makes explicit use of the action $\tau$. There is no way to define a cocycle without mention of a chosen group action of $F$ on $H$.

What you are missing, I think, is that this action becomes conjugation in the group extension. When we write $1 \to H \to G \to F \to 1$, it is implicit that the group action of $F$ on $H$ we are considering is conjugation inside $G$. If $t_\sigma H$ is identified with an element $\sigma \in F$, then we would expect $$\sigma \cdot h = t_\sigma h t_\sigma^{-1}$$

Conversely, when we have two groups $F$ and $H$, and we want to find a $G$ such that $1 \to H \to G \to F \to 1$ is exact, part of the data required in finding such a $G$ is a fixed action of $F$ on $H$, which will become the conjugation action in $G$.

This is all to say that you are perfectly able to assume that the action of $F$ on $H$ is conjugation. There is no other relevant action. If I have misunderstood your question, feel free to comment and clarify. I'll be happy to update my answer.

---

I hope this helps ^_^