# Given surjective Homomorphism $f:A \to B$ then $A \cong \ker(f) × B$

Given surjective Homomorphism $$f:A \to B$$ when $$B$$ is a free abelian group.

Prove: $$A \cong \ker(f) × B$$

I have seen this claim in the following link:

Abelian group admitting a surjective homomorphism onto an infinite cyclic group

But couldn't find any proof online. I have tried to prove it using the following claim but I don't think the same claim works for general $$\mathbb{Z}^n$$ https://yutsumura.com/surjective-group-homomorphism-to-z-and-direct-product-of-abelian-groups/

Please note that I haven't learned exact sequences yet.

• Let $\{b_i : i \in I\}$ be a free basis of $B$. For each $i \in I$ choose $a_i \in A$ with $f(a_i) = b_i$, and let $C$ be the subgroup of $A$ generated by the $a_i$. Prove that $C \cong B$ and $A = \ker f \times C$. Commented Feb 18, 2020 at 11:53
• @Derek Holt This is exactly what I was trying to do. How ever I can't see why $C \cong B$? Commented Feb 18, 2020 at 11:55
• Clearly $f(C) = B$, and so $B \cong C/(C \cap \ker f)$. So you need to prove that $C \cap \ker f = \{0\}$. This is where you use the fact that $B$ is free abelian. Commented Feb 18, 2020 at 12:28

Well, you say that you didn't learn exact sequences yet, but what exactly stops you from doing that? This tool is not difficult at all and reduces lots of boilerplate.

So you have a short exact sequence of abelian groups:

$$0\to \ker(f)\to A\xrightarrow{f} B\to 0$$

with $$B$$ free abelian. Let $$\{b_i\}$$ be a basis of $$B$$. For each $$i$$ choose $$a_i\in A$$ such that $$f(a_i)=b_i$$ and define $$g:B\to A$$ by $$g(b_i)=a_i$$. Since $$B$$ is free abelian then $$g$$ is a well defined homomorphism. With that you can easily check that the sequence splits and so we can apply the splitting lemma to conclude that $$A\simeq\ker(f)\times B$$.

Side note. This is a special case of the more general fact that if $$R$$ is a ring then a $$R$$-module $$P$$ is projective if and only if every sequence of the form

$$0\to A\to B\to P\to 0$$

splits. It just happens that the category of abelian groups is the same as the category of $$\mathbb{Z}$$-modules and projective modules coincide with free abelian groups.