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.
Best Answer
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.