Q:

Why is the sequence exact?

Bredon states:

For $A \subset X$, the sequence
$$ 0 \rightarrow \Delta_{*}(A) \otimes G \rightarrow \Delta_{*}(X) \otimes G \rightarrow \Delta_{*}(X,A) \otimes G \rightarrow 0$$
is exact because of the splitting map $\Delta_{*}(X,A) \rightarrow \Delta_{*}(X)$

I think that everything can be proved directly (without using the splitting map), except for the exactness at the center. Hence, I think we must use the splitting map to prove that, but I can't pinpoint how. How to proceed?

A:

More generally, if $F:Ab\to Ab$ is any functor which preserves addition of maps, then it sends split exact sequences to split exact sequences.  This follows from the following theorem:

Theorem Let $A\stackrel{i}{\to} B\stackrel{p}{\to} C$ be a pair of maps of abelian groups.  Then the following are equivalent:

There exist maps $q:B\to A$ and $j:C\to B$ such that $qi=1$, $pj=1$, and $iq+jp=1$.

The sequence $0\to A\to B\to C\to 0$ is exact, and there exists a map $q:B\to A$ such that $qi=1$.

The sequence $0\to A\to B\to C\to 0$ is exact, and there exists a map $j:C\to B$ such that $pj=1$.

There exists an isomorphism $f:B\to A\oplus C$ such that $fi:A\to A\oplus C$ is the inclusion map $a\mapsto (a,0)$ and $pf^{-1}:A\oplus C\to C$ is the projection map $(a,c)\mapsto c$.

Proof: Let us prove $4\Rightarrow 3\Rightarrow 2\Rightarrow 1\Rightarrow 4$.
($4\Rightarrow 3$): It is easy to see the sequence is exact.  To get $j$, compose the inclusion $C\to A\oplus C$ with $f^{-1}$.
($3\Rightarrow 2$): For $b\in B$, let $\tilde{q}(b)=b-j(p(b))$.  Then $p(\tilde{q}(b))=p(b)-p(j(p(b)))=p(b)-p(b)=0$, so $\tilde{q}(b)\in \ker(p)$.  Since $0\to A\to B\to C$ is exact, there is a unique element $q(b)\in A$ such that $i(q(b))=\tilde{q}(b)$, and $q:B\to A$ is a homomorphism.  Furthermore, if $a\in A$, then $\tilde{q}(i(a))=i(a)-j(p(i(a)))=i(a)$ since $pi=0$, so $qi=1$.
($2\Rightarrow 1$): Given $c\in C$, choose a $b\in B$ such that $p(b)=c$, and let $j(c)=b-i(q(b))$.  This does not depend on the choice of $b$ since any other choice is of the form $b+i(a)$ for some $a\in A$, and $i(q(i(a)))=i(a)$.  This defines a homomorphism $j:C\to B$, and $p(j(c))=p(b)-p(i(q(b)))=p(b)=c$, so $pj=1$.  It is also immediate from the definition of $j$ that $iq+jp=1$.
($1\Rightarrow 4$):  First, note that $pi=p(iq+jp)i=piqi+pjpi=pi+pi$, so $pi=0$.  Similarly, $qj=0$.  Now define $f:B\to A\oplus C$ by $f(b)=(q(b),p(b))$ and $g:A\oplus C\to B$ by $g(a,c)=i(a)+j(c)$.  The fact that $f$ and $g$ are inverse and that $fi$ and $pg$ have the desired form follows immediately from the identities $qi=1$, $pj=1$, $pi=0$, $qj=0$, and $iq+jp=1$.
If these conditions hold, the sequence $0\to A\to B\to C\to 0$ is called split exact.  In particular, condition (1) is manifestly preserved by any functor $F:Ab\to Ab$ which preserves addition of maps, so such a functor preserves all the other conditions as well.  In your case, $F$ is the functor $-\otimes G$, and you have a sequence that you know satisfies (3).  It follows that the sequence obtained by applying $F$ still satisfies (3), and in particular is exact.

