# Existence of abelian group extension relative to group homomorphism

Let $$f: A \to B\$$ be an abelian group homomorphism. Are there abelian groups $$G,\ H,\ K$$ such that $$K \subseteq H \subseteq G$$ and the map $$\pi \circ i: H \to G/K$$ which is the composition of projection and inclusion is isomorphic to $$f$$? By isomorphic to $$f\$$I mean there exist isomorphisms $$\tau: H \to A\$$ and $$\sigma: G/K \to B$$ such that $$f = \sigma \circ \pi \circ i \circ \tau^{-1}$$.

I think it is a quite natural question, since the case where $$f$$ is epimorphism is a consequence of the fundamental homomorphism theorem. However, I can't prove the existence nor the uniqueness of the group extension. I'm not familiar at all with the general theory of group extension, but maybe the case where $$A,\ B\$$are abelian could be easier.

The question has been at MSE (link) for two days, but there was no answer.

• As I said in my comment on the MSE question, I believe that the abelian group case is harder rather than easier, because there are easy examples that show that the answer is no (in general) in the nonabelian case. Nov 14, 2021 at 11:20

Let me identify $$A$$ with $$H$$. Clearly, $$K$$ can be identified with $$\ker(f)$$. Let $$f(A)$$ be the image of $$A$$ in $$B$$. Then $$A$$ is an abelian extension of $$f(A)$$ by $$K$$. If I understand correctly, you are asking whether given an inclusion $$f(A)\hookrightarrow B$$, it is possible to find an extension of $$B$$ by $$K$$, extending the given extension of $$f(A)$$. Equivalently, you are asking if it is possible to construct a diagram of the following form, where the rows are short exact, and all the vertical homomorphisms are monomorphisms: $$\require{AMScd} \begin{CD} K @>>> A @>f>> f(A)\\ @V=VV @VVV @VVV \\ K @>>> G @>>> B \end{CD}$$ The set of isomorphism classes of abelian extension of $$B$$ by $$K$$ is in bijective correspondence with $$\operatorname{Ext}(B, K)$$, where Ext is taken in the category of abelian groups. The inclusion of groups $$f(A)\hookrightarrow B$$ induces a surjective homomorphism $$\operatorname{Ext}(B, K)\twoheadrightarrow \operatorname{Ext}(f(A), K).$$ Because of the surjectivity, a group extension of $$f(A)$$ can always be extended to a group extension of $$B$$.
The reason that the homomorphism is surjective is that the cokernel would be a subgroup of $$\operatorname{Ext}^2(B/f(A), K)$$, but the category of abelian groups has projective dimension one, so Ext$$^2$$ is always zero.
On the other hand, the homomorphism of ext groups is not always injective. The kernel is a quotient of $$\operatorname{Ext}(B/f(A), K)$$. So the lift is not unique.
You can construct an extension of $$B$$ explicitly as follows. It is a standard result of homological algebra that you can construct a map of free resolutions of the inclusion $$f(A)\hookrightarrow B$$. $$\require{AMScd} \begin{CD} 0@>>> R_A @>>> F_A @>>> f(A)\\ @. @VVV @VVV @VVV \\ 0 @>>> R_B @>>> F_B @>>> B \end{CD}$$ Such that the homomorphisms $$R_A\to R_B$$ and $$F_A\to F_B$$ are split injections. An extension of $$f(A)$$ by $$K$$ induces a homomorphism $$R_A\to K$$. Use the splitting to get a homomorphism $$R_B\to K$$. Now you have a surjective homomorphism $$F_B \oplus_{R_B} K \twoheadrightarrow B$$, whose kernel is $$K$$. This is the desired extension. In your notation, $$G=F_B \oplus_{R_B} K$$.