# Sequence splits if and only if $\varphi$ has right inverse

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.

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.