Left inverse of a homomorphism in abelian groups

Question

An injective homomorphism $f: A \rightarrow B$, where $A, B$ are abelian groups has a left inverse iff $f(A)$ is a direct summand of $B$. It's clear to me that the image has to be a normal subgroup of $G$. Isn't this a necessary condition for an inverse homomorphism? Why is the direct summand condition needed, where it comes from?

Answer 1

If $f(A)$ is a summand, i.e. $B=f(A)\times N$, then the map $(f(a),n)\mapsto a$ you should check is well-defined (because $f$ is one-to-one), is a homomorphism, and is a left inverse of $f$.

Conversely, if there is a left inverse $\ell:B\to A$ for which $\ell\circ f=\mathrm{id}_A$, then letting $N$ be the kernel of $\ell$ you should check that $f(A)\cap N$ is trivial, which combined with the fact everything is abelian (implying they are both normal subgroups) says $B=f(A)\times N$ is an internal direct product.

Thus, existence of left inverse is equivalent to image being a direct summand.

Note if $f(A)$ is not a direct summand then the existence can fail. The embedding $\mathbb{Z}_2\to\mathbb{Z}_4$ is one-to-one but it does not admit a left inverse nor is the image a direct summand. (But there is a projection $\mathbb{Z}_4\to\mathbb{Z}_2$, of course, it is just not a left inverse of the embedding $\mathbb{Z}_2\to\mathbb{Z}_4$; can you check?)