Ytukyg

I need some help to understand the proof of the following theorem:

Theorem If $Asubset X$ is a retract of $X$ then, $$H_n(X)simeq H_n(A)oplus H_n(X, A),$$ all $ngeq 0$.

Proof. Let $r:Xlongrightarrow A$ be a retraction. Since $rcirc imath=id_A$ it follows $r_*circ imath_*=id_{H_n(A)}$ hence $imath_*$ is injective. Consider the exact homology sequence of the pair $(X, A)$: $$ldotslongrightarrow H_n(A)stackrel{imath_*}{longrightarrow} H_n(X)stackrel{jmath_*}{longrightarrow} H_n(X, A)stackrel{partial}{longrightarrow}H_{n-1}(A)longrightarrowldots$$ Since $imath_*$ is injective we find $textrm{ker}(imath_*)=0=textrm{im}(partial)$. But this says $jmath_*:H_n(X)longrightarrow H_n(X, A)$ is onto for all $ngeq 0$. In other words, for all $ngeq 0$, we have a short exact sequence $$0longrightarrow H_n(A)stackrel{imath_*}{longrightarrow} H_n(X)stackrel{jmath_*}{longrightarrow}H_n(X, A)longrightarrow 0,$$ which splits. Therefore, for all $ngeq 0$, $H_n(X)simeq H_n(A)oplus H_n(X, A)$.

Question: Where did the $0's$ in $0longrightarrow H_n(A)stackrel{imath_*}{longrightarrow} H_n(X)stackrel{jmath_*}{longrightarrow}H_n(X, A)longrightarrow 0$ come from?

We have ${rm im}(partial)=0$, so $partial=0$ as map $H_n(X,A)to H_{n-1}(A)$, so it factors through the zero object: $H_n(X,A)to 0to H_{n-1}(A)$. Extending the long exact sequence with these $0$'s will keep the exactness.

Anyway, injectivity of a map $f$ is equivalent to $ker f=0$, i.e. $0to A overset{f}to B$ is exact, and dually, surjectivity is equivalent to $Ato Bto 0$ being exact.

The reason why one can obtain a short exact sequence $0 to H_n(A) xrightarrow{i_*} H_n(X) xrightarrow{j_*} H_n(X, A) to 0$ in the proof is expressed generally by the result below

Result: Consider an exact sequence of abelian groups $$dots to C_{n+1} to A_n xrightarrow{i_n} B_n xrightarrow{p_n} C_n to A_{n-1} xrightarrow{i_{n-1}} B_{n-1} xrightarrow{p_{n-1}} C_{n-1} to dots$$ in which every third map $i_n$ is injective. Then the sequence $$0 to A_n xrightarrow{i_n} B_n xrightarrow{p_n} C_n to 0$$ is exact for all $n$

Now in the proof supplied in the question, we have the following long exact sequence (of abelian groups)

$$dots H_{n+1}(X, A) xrightarrow{partial} H_n(A) xrightarrow{i_*} H_n(X) xrightarrow{j_*} H_n(X, A) xrightarrow{partial} H_{n-1}(A) xrightarrow{i_*} H_{n-1}(X) xrightarrow{j_*} H_{n-1}(X, A) todots$$

and every third map $i_*$ is injective, so by the result above we obtain a short exact sequence $0 to H_n(A) xrightarrow{i_*} H_n(X) xrightarrow{j_*} H_n(X, A) to 0$ as desired.

This result quoted above is Exercise 5.14 (i) in Introduction to Algebriac Topology by J. Rotman. It is not too difficult to prove this result.