Proving the Splitting Lemma on pg.147 in AT.

Question

The Lemma and a sketch of its proof are given below:

My questions are:

1- why the diagram that must be commutative looks specifically like this? My professor actually drew a diagram without the zeros on both ends and with the arrow from $C$ to $B$ reversed and also the arrow from $A \oplus C$ to $B$ reversed and also the arrow from $A \oplus C$ to $C$ reversed?Also, is this the pushout diagram?

2- Could anyone show me the details of proving that (b) implies (c) and how can we use the Five Lemma?

EDIT:

How are the exact sequences in the short five lemma will look like and why? I got a hint that they will look like $0 \rightarrow A \rightarrow A\oplus C \rightarrow C \rightarrow 0$

And the second line will look $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$, but this trivially will give us the desired isomorphism without even using the homomorphism given in (b)

Also, I got a hint that the diagram commutes just because of the direct sum which I do not understand.

Answer 1

For your first question, it sounds like you were given a diagram that looked like this : $$\require{AMScd} \begin{CD} A @>i_A>> A\oplus C @<i_C<< C \\ @V1_AVV @V{\varphi}VV @VV1_CV \\ A @>>i> B @<<s< C\end{CD}$$ (I can't do a triangle here, so I've added vertical maps on the sides, they are just identities). If this is what you've seen, this is to indicate that the map $\varphi$ is defined thanks to the universal property of the direct sum $A\oplus C$ : it is the only map such that $\varphi i_A=i$ and $\varphi i_C=s$. Since you can write $i_A(a)=(a,0)$ and $i_C(c)=(0,c)$, you have $$\varphi(a,c)=\varphi(a,0)+\varphi(0,c)=i(a)+s(c)$$ so it is the same map as in Hatcher's book. Now that we have constructed the map, we want to show that it is an isomorphism. For that you can just use the short five lemma on the diagram $$\require{AMScd} \begin{CD} 0 @>>> A @>i_A>> A\oplus C @>{p_C}>> C @>>> 0 \\ & @V1_AVV @V{\varphi}VV @VV1_CV & \\ 0 @>>> A @>>i> B @>>j> C @>>> 0.\end{CD}$$ Indeed the two rows are exact sequences, the left square commutes by definition of $\varphi$, and you can easily check that the right square also commutes. Since the vertical sides are isomorphisms (even identities) $\varphi$ must also be an isomorphism.

Answer 2

Arnaud D. put a nice answer for your question 2, so I'll just clarify some things about your question 1. I'll use the same notation as in that answer.

We want a good way to define a "split exact sequence." These are the nicest kinds of short exact sequences because they're very easy to understand: The map $i_A: A \to A \oplus C$ sends $a \mapsto (a, 0)$ and the map $p_C: A \oplus C \to C$ sends $(a,c)\mapsto c$. So we want a way to describe short exact sequences that also behave this way, so that we have an isomorphism of the middle term onto the direct sum of the adjacent terms. This motivates the definition that a short exact sequence $0 \to A \to B \to C \to 0$ is split exact if there is a commutative diagram

$$\require{AMScd} \begin{CD} 0 @>>> A @>i_A>> A\oplus C @>{p_C}>> C @>>> 0 \\ & @V1_AVV @V{\varphi}VV @VV1_CV & \\ 0 @>>> A @>>i> B @>>j> C @>>> 0\end{CD}$$

where $\varphi$ is an isomorphism.

So this is why the commutative diagram has to look exactly like the one you've drawn (note that because the vertical maps $A \to A$ and $C \to C$ are both the identity, the commutative diagram in your picture is the same as this one). Thus the splitting lemma really says that a short exact sequence is split exact (i.e., the words in (c)) iff there is a left-inverse to $i$ iff there is a right-inverse to $j$.

I don't quite know how to interpret the rest of your question (1). The diagram in the splitting lemma is not quite the pushout diagram in the category of abelian groups; that pushout is described in the examples of a pushout on Wikipedia. But what your professor said is equivalent to the splitting lemma, as follows.

Provided your sequence is split exact, you can show that B is (isomorphic to) the pullback of the diagram $C \to C \leftarrow A\oplus C$ (a hint for this: use the map $s$ and the fact that $\varphi$ is an isomorphism to show that $B$ satisfies the universal property). Similarly, you can show that if the sequence is split exact, then $A \oplus C$ is the pushout of the diagram $A \leftarrow A \to B$. The maps I've not labeled are the ones you already have (respectively $1_C, p_C, i_A, i$).

Similarly, the converses hold (pushout implies existence of a left-inverse for $i: B \to A$). Thus we have the following: A short exact sequence is split exact iff $i$ has a left-inverse iff $j$ has a right-inverse iff $A \oplus C$ is the pushout of $A \leftarrow A \xrightarrow{i} B$ iff $B$ with map $j: B\to C$ is the pullback of $A \oplus C \to C \leftarrow C$.