Let $\varphi:M\rightarrow N$ be an $R$-module homomorphism. Prove that the short exact sequence
$$0 \rightarrow\text{ker}\varphi \rightarrow M \xrightarrow{\varphi}N\rightarrow 0$$
splits if and only if $\varphi$ has a right inverse.
Attempt:($\implies$) Assume the sequence splits. Then it is isomorphic to a sequence of the form
$$0\rightarrow \text{ker}\varphi \xrightarrow{\iota} \text{ker}\varphi \oplus N \xrightarrow{\pi} N \rightarrow0$$
Where $\iota$ and $\pi$ are the inclusion and projection maps. Thus $\iota:N \rightarrow \text{ker} \varphi \oplus N$ is a right inverse map to $\pi$, since
$$\pi \circ \iota(a)=\pi(0,a)=a$$
Since we can identify $\varphi$ with $\pi$, we can identify $\iota$ as the right inverse map to $\varphi$.
Comment: This last sentence seems a little imprecise. How can I be clearer? It doesn't seem right to say the maps $\varphi$ and $\pi$ are isomorphic. So what are they?
($\Longleftarrow$)
This problem is similar from an example in Aluffi's chapter 0, so I will try and adapt the proof. Suppose $\varphi$ has a right inverse $\psi:N \rightarrow M$. To show the sequence splits, I want to show that $M$ is isomorphic to $\text{ker}\varphi \oplus N$. But it seems that it would be easier to show $M \cong \text{coker} \psi\oplus N$(Assuming $\text{coker} \psi \cong \text{ker} \varphi$ which I am unable to verify).I will try to construct such an isomorphism. I was thinking $\alpha:\text{coker}\psi \oplus N \rightarrow M$ given by $(n+\psi(N),p) \mapsto n+\psi(p)$. This map does not seem right to me. Should I even be trying to show $M \cong \text{coker} \psi \oplus N$? However, even if this is the correct map, I am struggling to come up with an inverse. How should I proceed to do this to finish the problem?
In the left inverse case with short exact sequence
$$0 \rightarrow M\xrightarrow{\varphi}N \rightarrow \text{coker}\varphi \rightarrow 0$$
the author shows $N \cong M \oplus \text{ker} \psi$ where $\psi$ is the right inverse to $\varphi$, by constructing the map $M \oplus \text{ker} \varphi \rightarrow N$ given by $(m,k) \mapsto \varphi(m)+k$ and gives an inverse $n \mapsto (\psi(n),n-\varphi \psi(n))$ to show the isomorphism. However, I have no idea how he constructs the inverse.
Best Answer
To address your first comment: you can be less imprecise there by being more explicit about the isomorphism. In particular, if the sequence $$ 0 \to \ker \varphi \xrightarrow{i} M \xrightarrow{\varphi} N \to 0 $$ splits you have an isomorphism $f : M \to \ker \varphi \oplus N$ such that $f \circ i = \iota_1$ and $\pi_2 \circ f = \varphi$, where $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$ and $\pi_2 : \ker \varphi \oplus N \to N$ are the corresponding inclusion and projection.
Then you have for $\iota_2 : N \to \ker \varphi \oplus N$ that $\pi_2 \circ i_2 = \mathrm{Id}_N$ and you define $r : N \to M$ by $r = f^{-1} \circ \iota_2$. Then $\varphi \circ r = \varphi \circ f^{-1} \circ \iota_2 = \pi_2 \circ \iota_2 = \mathrm{Id}_N$ so you have a right inverse.
For the converse, assume you have a right inverse $r : N \to M$. Note that a right inverse is injective, since $r(x) = r(y)$ implies $x = \varphi(r(x)) = \varphi(r(y)) = y$. Thus, we can identify $N$ with the submodule $\operatorname{im} r$ of $M$.
Our goal is to show that $M = \ker \varphi \oplus N$.
Now, to see how this can be done, let $m \in M$ . Then $n = \varphi(m)$ is an element of $N$ and since $r$ is a right inverse, for $m' = r(n)$ you have that $$\varphi(m - m') = \varphi(m) - (\varphi \circ r \circ \varphi)(m) = \varphi(m) - \varphi(m) = 0.$$
Thus $m - m' = i(k)$ for some unique $k \in \ker \varphi$ and we can write $m = i(k) + r(n)$.
Define $f : M \to \ker \varphi \oplus N$ by this decomposition, i.e. $$ f(m) = (k, \varphi(m)). $$
You can verify this is an isomorphism with inverse $g : \ker \varphi \oplus N$ given by $g(k, n) = i(k) + r(n)$
Now, since $\varphi \circ i = 0$ you have $(f \circ i)(k) = (k, 0)$, i.e. is the inclusion $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$. Similarly for $\pi_2 : \ker \varphi \oplus N \to N$ you have $(\pi_2 \circ f)(m) = \varphi(m)$, so $\pi_2 \circ f = \varphi$.
Thus the exact sequence splits, with $f$ giving the required isomorphism between exact sequences.