Let $0\to N\to M\to P\to 0$ be a split exact sequence. Show that $\varphi:N\oplus P\to M$ given by $\varphi(n,p)=f(n)+h_1(p)$ is an isomorphism.

Question

I'm working on the following problem:

Consider a short exact sequence of $R$-modules: $$0\to N\xrightarrow{f} M\xrightarrow{g} P\to 0. $$ Suppose that $h_1:P\to M$ is an $R$-map such that $g\circ h_1=\text{id}_P$. Define a map $\varphi:N\oplus P\to M$ by $\varphi(n,p)=f(n)+h_1(p)$. Prove that $\varphi$ is an $R$-isomorphism.

So far I was able to prove that $\varphi$ is an injective $R$-map. I'm just having trouble proving it's surjective. If we take $m\in M$, I know that if $m\in im(f)$ then $m=f(n)$ for some $n\in N$, so in particular, $\varphi(n,0)=f(n)=m$. But what if $m\notin im(f)$? That's where I'm stuck. I've thought about using $g(m)$ in the second entry since it is in $P$, and am not seeing the clever algebra trick I'm sure there is for writing something in the first entry so that it simplifies to $m$. Any hints?

Answer 1

Consider $m\in M$ and then let $p=g(m)$. Then

$$g(m-h_1(p))=g(m) - p=0$$

and since the sequence is exact there is $n\in N$ such that

$$f(n)=m-h_1(p)$$

That's it.