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

enter image description here

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

$0 \to M \stackrel{\phi_1}{\rightarrow} N \to coker \ \phi \to 0$


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 $M'_2 \cong \frac{N}{im \ \phi} \cong coker \ \phi$


2 Answers 2


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 $x\in M$, $$ \beta\phi(x)=(\psi\phi(x),\pi\phi(x))=(x,0)=f(x) $$

  • $\begingroup$ Can you please elaborate on "it is implicit in the definition that the maps in the bottom row are the canonical ones, so there exists $h" M' \oplus C' \to M'$ such that $hf = 1_{M'}$? How do you know it's necessarily that $f(m') = (m',0)$ and $g(m',c') = c'$? Unfortunately, I don't see it. $\endgroup$
    – Jxt921
    Commented Jun 10, 2016 at 18:49
  • $\begingroup$ @Jxt921 The maps must be those, or the result would be false. It's not necessarily true that an exact sequence $0\to A\to A\oplus B\to B\to 0$ is split. The “canonical” exact sequence is. $\endgroup$
    – egreg
    Commented Jun 10, 2016 at 19:45
  • $\begingroup$ Could you show me how do we deduce that? If it's split, than we have $\beta \phi = f \alpha$. Still, an exact definition for $f$ relies on $\beta, \phi$ and $\alpha$. I just don't see how we get from the diagram you drew to the formula for $f(m') = (m', 0)$. Or we have from the diagram $\gamma(\phi(m) + im \ \phi) = g(f(\alpha(m)))$. Still doesn't say much about $f$ and $g$ to claim they are canonical. $\endgroup$
    – Jxt921
    Commented Jun 10, 2016 at 20:21
  • 1
    $\begingroup$ @Jxt921 $f$ is defined as $f(x)=(x,0)$. It is the canonical injection in the direct sum. $\endgroup$
    – egreg
    Commented Jun 10, 2016 at 20:23
  • 1
    $\begingroup$ @Jxt921 No, not some $f$ and $g$; the map $f$ is defined by $f(x)=(x,0)$ and $g$ by $g(x,y)=y$. This is the implicit assumption in the book definition. Otherwise you have no hope to prove the result. $\endgroup$
    – egreg
    Commented Jun 10, 2016 at 20:31

EDIT: Based on your comment I changed the answer slightly to solve the confusion...

The point is we have an injective map $\phi_1: M\to N$ and we are looking for a map (a retraction) $\rho: N\to M$ such that $\rho\phi_1 = \mathrm{id}_M$. Let $C=\mathrm{Coker}\phi_1$. The exact sequence $0\to M\xrightarrow{\phi_1} N\xrightarrow{\kappa} C\to 0$ is split (where $\kappa$ is the quotient map). So there are $R$-modules $M'_1, M_2'$ such that the diagram you drew is commutative with all the vertical maps isomorphisms. Let us put some names on these isomorphisms $\alpha: M\xrightarrow{\sim} M'_1$, $\beta: C\xrightarrow{\sim} M'_2$ and $\psi: N\xrightarrow{\sim} M'_1\oplus M'_2$.

(This part is new) First of all note that you also have an exact sequence $0\to M\xrightarrow{\imath} M\oplus C\xrightarrow{\pi}C\to 0 $ in addition to $0\to M'_1\xrightarrow{\imath'} M_1'\oplus M_2'\xrightarrow{\pi'}M_2'\to 0 $. Clearly $$ \imath' \circ \alpha = (\alpha\oplus\beta)\circ \imath, \quad \beta\circ\pi = \pi'\circ (\alpha\oplus\beta) $$ This means you can without loss of generality assume that $M'_1=M$, $\alpha=\mathrm{id}_{M}$, $M'_2=C$, $\beta=\mathrm{id}_C$ since 0\to M\xrightarrow{\imath} M\oplus C\xrightarrow{\pi}C\to 0$ also satisfies the conditions you listed in the definition of split exact sequence.

Consider the map $\rho:N\xrightarrow{\psi}M\oplus C\xrightarrow{\pi_M}M$ where the second map is the projection to $M$. The claim is $\rho\phi_1=\mathrm{id}_M$. Note that $\mathrm{Im}(\phi_1)\simeq M$ because $\phi_1$ is injective. Moreover this isomorphism is exactly $\psi$ itself. So take $x\in M$, then $\psi\phi_1(x)=(x,0)$. Hence $\rho\phi_1$ is indeed identity. It is in this sense that $\phi_1$ can be identified with the embedding $M\to M\oplus C$.

  • $\begingroup$ That's what I don't understand: why we have an isomorphism $\psi: N \to M'_1 \oplus M'_2$? We know there are $M'_1 \cong M, M'_2 \cong coker \ \phi$ and $M'_1 \oplus M'_2 \cong N$ but we DON'T know that these isomorphisms are indentities and $M = M'_1$ and $coker \ \phi = M'_2$. The definition of split exact sequence says "isomorphic" and not "equal". I know sometimes structures are considered to be the same up to isomorphism, but I don't see it as the case, since we need to construct specific functions, like projection. $\endgroup$
    – Jxt921
    Commented Jun 9, 2016 at 21:38
  • $\begingroup$ I added a paragraph which should resolve your confusion: The definition of split exact sequence says "there exists" such and such... I explained that you can choose the downstairs exact sequence such that $M_1'=M$ and $M'_2=C$ but the last one $M'_1\oplus M_2'\simeq N$ is just an isomorphism. $\endgroup$
    – Hamed
    Commented Jun 9, 2016 at 22:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .