Confusion about equivalent definition of short exact sequence split

Question

I am trying to prove this exercise in Rotman's Advanced Algebra. The objects here are all $R$-modules and the arrows are $R$-maps.

Exercise 7.17: Prove that a short exact sequence $$ 0 \to A \to^i B \to^p C \to 0 $$ splits if and only if there exists $q : B \to A$ with $qi = 1_A$.

The definition of split he is using is:

Definition: A short exact sequence $$ 0 \to A \to^i B \to^p C \to 0 $$ is split if there exists a map $j : C \to B$ with $pj = 1_C$.

I am stuck trying to show this. Intuitively what I need to do is to 'reverse' $p$ knowing that $q$ is a $R$-map. So (guessing here) I want to define for all $c \in C$, $f(c)$ to be $p^{-1}(c) - iq p^{-1} (c)$ where $p^{-1}(c)$ is any choice of pre-image. But I am not sure how to proceed from here. So my first question is: Is this correct and how do I continue / what should I do instead?

As I was searching this site for an answer, I came across some other questions that made me confused. They are:

1. Short Split Exact Sequence Theorem

This question is about 3 equivalent conditions for a short exact sequence to split. For condition (2), my understanding is that (using the same symbols except he uses $q$ where I have used $p$) $D = B / i(A) = B/ker(q) = C$. Hence, condition (2) says that the sequence splits if and only if $B = A \oplus C$. But isn't this false in general? In Rotman's book as well as elsewhere on this site there are examples of this.

2. https://math.stackexchange.com/a/2823560/577979

Similarly in this question, the accepted answer seems to be using the same reasoning as in the above question.

Could you clarify this for me too? Thank you very much!

Answer 1

The answers are correct.

There is, however, a subtlety in the meaning of "$B=A\oplus C$". It should be read e.g. as "$B \simeq A \oplus C,$ and under this identification, the map $A \rightarrow B$ becomes inclusion to the first component and $B \rightarrow C$ becomes projection onto the second component". The counterexamples to $B=A \oplus C$ implying split are considering some different maps than these into the non-split short exact sequence.

(Also note that e.g. in the category of (not necessarily abelian) groups the splitting on the left is not equivalent to splitting on the right: splitting on the left implies that the middle term is the direct product of the left and right term (again, in the above sense) while splitting on the right gives only semidirect product.)

Answer 2

Proof:

$\Leftarrow )$

Let $\varphi : B \to A\times C$

$\varphi(b) = (q(b), p(b))$

Then $\varphi$ is an isomorphims, well, $q,p$ are surjectives, and $\varphi(b) = 0 \iff q(b) = 0 = p(b)$

So $b \in Ker(p) = Im(i) \implies i(a) = b$ for some $a \in A$, then $0 = q(b) = q(i(a)) = a \implies 0 = i(a) = b$

Then $\varphi$ is injective.

Now we have that $B \simeq A\times C$

Let $i_{2} :C \to A\times C$, $i_{2}(c) = (0,c)$

Let $\pi_{2}:A \times C \to C$, $\pi_{2}(a,c) = c$

Thus $\pi_{2} i_{2} = 1_{C}$

Note that $p = \pi_{2} \varphi$, then if we take $j = \varphi^{-1} i_{2}$ we have that $pj= \pi_{2} \varphi \varphi^{-1} i_{2} = \pi_{2} i_{2} = 1_{C}$ then it holds.

$\Rightarrow )$

We have that there exist $j:C \to B$ such that $pj = 1_{C}$

Let $\varphi: A\times C \to B$, $\varphi(a,c) = i(a) + j(c)$

Let $b \in B$ then $p(jp(b)) = pj(p(b)) = 1_{C}(p(b)) = p(b) \implies 0 = p(b - jp(b)) \implies b - jp(b) \in Ker(p)$

Then, as $Im(i) = Ker(p)$ we have that $i(a) = b - jp(b) \implies i(a) + j(p(b)) = b$

So $\varphi(a,p(b)) = b$ then $\varphi$ is surjective.

Now, $\varphi(a,c) = 0 \iff i(a) + j(c) = 0 \iff i(a) = j(-c) \iff j(-c) \in Ker(p)$

$ \iff 0 = p(j(-c)) = 1_{C}(-c) = -c \iff c = 0$

Then $i(a) = 0$, and $i$ is injective $\implies a = 0$

Thus $\varphi$ is injective, then is an isomorphims

Let $\pi_{1}: A \times C \to A$, $\pi_{1}(a,c) = a$

Let $i_{1}:A \to A \times C$, $i_{1}(a) = (a,0)$

Note that $i = \varphi i_{1}$

If we take $q = \pi_{1} \varphi^{-1}$, we have that $qi = \pi_{1} \varphi^{-1} \varphi i_{1} = \pi_{1} i_{1} = 1_{A}$

And we are done.