I was attempting to prove that for an exact sequence $A_1\overset{\varphi_1}{\to}A_2\overset{\varphi_2}{\to}A_3...$, the sequence $...\hom(A_3,\mathbb{R})\overset{\varphi_2^*}{\to}\hom(A_2,\mathbb{R})\overset{\varphi_1^*}{\to}\hom(A_1,\mathbb{R})$ is also exact, where $\varphi_k^*(f)$ is the homomorphism given by $\varphi_k^*(f)(x)=f(\varphi_k(x))$.

When showing that if $\varphi_k^*(f)=0$, then $f=\varphi_{k+1}^*(g)$, I simply let $g$ be the homomorphism defined on the image of $\varphi_{k+1}$ as $g(x)=f(y)$ if $x=\varphi_{k+1}(y)$. This is well defined because if $z$ is such that $x=\varphi_{k+1}(z)$, then $\varphi_{k+1}(y-z)=0$, so $\varphi_k(w)=y-z$ and $f(y-z)=f(\varphi_k(w))=\varphi_k^*(f)(w)=0$.

Now my problem is to extend $g$ to all of $A_{k+2}$, not just on the image $\varphi_{k+1}$. All groups are abelian.

My question is now, given an abelian group $G$ and a subgroup $H$ and a homomorphism $f:H\to \mathbb{R}$ is it always possible to extend $f$ to a homomorphism on all of $G$? I have managed to prove it for infinite cyclic groups $G$, but nothing in general, any hints?


For finite cyclic groups it also holds since the only homomorphism from such a group to $\mathbb{R}$ is the trivial one.

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    $\begingroup$ Maybe the transfer map $Ver:G/[G,G]\to H/[H,H]$ may help? $\endgroup$ Aug 22, 2021 at 10:28
  • $\begingroup$ @DavidA.Craven Sorry, I meant infinite cyclic groups. $\endgroup$ Aug 22, 2021 at 13:11
  • $\begingroup$ @DavidA.Craven I haven't checked the infinite cyclic group now, but how is your first map a homomorphism? $\phi(x^2x^2)=0$ but $\phi(x^2)+\phi(x^2)=-2$. Am I making a mistake? $\endgroup$ Aug 22, 2021 at 13:18
  • $\begingroup$ Let $f:H\rightarrow\mathbb{R}$ and $g\notin H$. We can assume that $G=\langle g,H\rangle$. Then $G=\langle g\rangle+H$. (We consider $G$ to be an additive group). Then we consider two cases: 1) $\langle g\rangle\cap H=\{0\}$. In this case we take $f(g)=0$. 2) $kg\in H$ and $k$ a minimal positive integer with this property. Then each element $x\in G$ is uniquely represented as $x=lg+h$, where $0\leq l<k$ and $h\in H$. We can then put $f(g)=f(kg)/k$ and $f(x)=lf(g)+f(h)$. Wouldn't it work out that way. $\endgroup$
    – kabenyuk
    Aug 22, 2021 at 14:07
  • $\begingroup$ @kabenyuk that is fair, but how can you assume $G=\langle g,H\rangle$? $\endgroup$ Aug 22, 2021 at 14:51


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