Is the group cohomology for a profinite group always torsion?

Question

In his notes on group cohomology here, Bjorn Poonen claims that $H^i(G, A)$ is torsion when $G$ is profinite and $i>0$. why is the following not a counterexample? Take $G= \hat{\Bbb Z}$, and $A=\Bbb Z$ with the trivial $G$ action, so that $H^1(G,A)$ is just $Hom_{cont} (G, A)$. Then we have a the element $f: G \rightarrow A$ given by $f(1)=1$. This extends to a function on $G$ because $1$ is a topological generator. This should have infinite order in $H^1(G,A)$, unless I'm mistaken. If someone can help clear this up it would be much appreciated!

Answer 1

There is no such continuous homomorphism $f : \hat{\mathbb{Z}} \to \mathbb{Z}$. Indeed, since we are giving $\mathbb{Z}$ the discrete topology, $\ker f$ must be an open subgroup of $\hat{\mathbb{Z}}$. But open subgroups of profinite groups have finite index, so the image of $f : \hat{\mathbb{Z}} \to \mathbb{Z}$ must be finite, hence $f = 0$.

In particular, $H^1 (\hat{\mathbb{Z}}, \mathbb{Z}) = 0$.