I'm currently reading "Algebra: Chapter 0" by Paolo Aluffi.
Before I state my problem, I would like to give here the definition of split exact sequences from the book:
A short exact sequence
$0 \to M_1 \to N \to M_2 \to 0$
"splits" if there is a commutative diagram
in which the vertical maps are all isomorphisms.
Now, I'm having trouble with understandg a part of a given proof of the following proposition:
Let $\phi: M \to N$ be an injective $R$-module homomorphism. Then $\phi$ has a left-inverse(as a homomorphism, not a function) if and only if the sequence
$0 \to M \stackrel{\phi_1}{\rightarrow} N \to coker \ \phi \to 0$
splits.
Here is the proof from the book(I will state only the part I'm having trouble understanding with, that is, the "if" part of the statement):
If the sequence splits, then $\phi$ may be identified with the embedding of $M$ into a direct sum $M \oplus M'$, and the projection $M \oplus M' \to M$ gives a left-inverse of $\phi$.
Well, I'm a bit a lost here:
First of all, why $\phi$ can be identified with the embedding of $M \to M \oplus M'$? According to the definition above, it can be identidied with a surjective function $M'_1 \oplus M'_2$ where $M'_1 \cong M$ and $M'_2 \cong \frac{N}{im \ \phi} \cong coker \ \phi$
Best Answer
Suppose the sequence $0\to M\xrightarrow{\phi}N\to\operatorname{coker}\phi\to0$ splits. Then there exists the diagram $$\DeclareMathOperator{\coker}{coker}\require{AMScd} \begin{CD} 0 @>>> M @>\phi>> N @>\pi>> \coker\phi @>>> 0 \\ @. @V{\alpha}V\sim V @V\beta V\sim V @V\gamma V\sim V @. \\ 0 @>>> M' @>f>> M'\oplus C' @>g>> C' @>>> 0 \end{CD} $$ as specified in the definition.
Our task is to find $\psi\colon N\to M$ such that $\psi\phi=1_M$. It is implicit in the definition that the maps in the bottom row are the canonical ones, so there exists $h\colon M'\oplus C'$ such that $hf=1_{M'}$.
Now define $\psi=\alpha^{-1}h\beta$; then $$ \psi\phi=\alpha^{-1}h\beta\phi=\alpha^{-1}hf\alpha= \alpha^{-1}\alpha=1_M $$ The “identification” is exactly considering $f$; it seems that the author is trying to confuse the reader instead of applying the definition.
For proving the converse, you need to build the diagram by exploiting the existence of $\psi\colon N\to M$ with $\psi\phi=1_M$. The diagram is $$ \begin{CD} 0 @>>> M @>\phi>> N @>\pi>> \coker\phi @>>> 0 \\ @. @V{1_M}V\sim V @V\beta V\sim V @V1_{\coker\phi} V\sim V @. \\ 0 @>>> M @>f>> M\oplus\coker\phi @>g>> \coker\phi @>>> 0 \end{CD} $$ where $\beta(x)=(\psi(x),\pi(x))$. The diagram commutes because $g\beta=\pi$ and, for $x\in M$, $$ \beta\phi(x)=(\psi\phi(x),\pi\phi(x))=(x,0)=f(x) $$