0
$\begingroup$

Let $A$, $G$ be abelian groups, $R \to F \to A$ be the standard free resolution of $A$, $\phi:R \to G$ be a morphism and $N:=\text{coker}(-\phi \times i:R \to G \otimes F)$.

From MacLane - Homology i know that $\phi$ can be extended to $F$ iff the morphism $\pi \circ \iota_1:G \to N$ has a left inverse. On the other hand i know that $\text{Ext}(A,G)$ is trivial (i.e. $\phi$ is extendable) if $G$ is divisible, i.e. for all $g \in G$ and $n \in \mathbb Z/\text{{0}}$ there is an element $t\in G$ such that $g=nt$.

I'm trying now to show the existence of the inverse by using this divisibility, but i'm stuck... I think I have to show that $[g,0] \mapsto g$ is well-defined, so if $(g,0)$ ~ $(h,\sum \lambda_i r_i)$, then $g=h$, but i have no idea how this could possibly work...


Actually for now it would suffice for me to know why $Ext(A,G)$ is trivial, when G is divisible.

$\endgroup$
  • 2
    $\begingroup$ Hint: A divisible $\mathbb{Z}$-module is injective. $\endgroup$ – Tobias Kildetoft Sep 14 '13 at 18:55
  • $\begingroup$ Guess: $\pi\circ \iota_1$ is injective, so Im($\pi\circ \iota_1$)$\subset N$ is an injective $\mathbb Z$-module as well, so $N=$Im$(\pi\circ \iota_1)\oplus X$ which provides a left inverse $\endgroup$ – user83496 Sep 15 '13 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.