A module J that is a direct summand of every module containing J

Question

Suppose that J is a module with the property of being a direct summand of every module M containing J as a submodule. Now consider a short exact sequence $0\rightarrow J\rightarrow M\rightarrow N\rightarrow 0$, with $f:J\rightarrow M$ and $g:M\rightarrow N$ the morphisms of the sequence. This sequence splits? (an equivalent question is: is J injective?)

If $J\subset M$ and $f$ is the inclusion, then $M=J \oplus T$ for a submodule $T\subset M$, because of the property of J. Then $\pi:J\oplus T\rightarrow J$, the projection on $J$, is a retraction of $f$, then the sequence splits in this case.

If $J\not\subset M$, we can think that J is "included" in $M$ using the monomorfism $f$, since $J$ is isomorphic to a submodule $f(J)\subset M$. What happens in this case?

Answer 1

Being a direct summand of every supermodule is preserved under isomorphisms. This should immediately answer your question.

To see that, suppose $J\cong J'$ and $J'\leq M'$. Then you can build a module $M$ such that $J\leq M$ and $M\cong M'$ with the obvious diagram commuting. This is an easy exercise.

Answer 2

an equivalent question is: is J injective?

Of course it is, because it would be a summand of its injective hull ( hence equal to its injective hull.)