0
$\begingroup$

I need to prove the following statement:

Let $X$ be a deformation retract of $A$ by $r:X\to A$. Prove that $$H_q(X)\cong H_q(A)\oplus H_q(X,A)$$

Idea: As $r:X\to A$ is an homotopy equivalence, the induced map $H_q(r):H_q(X)\to H_q(A)$ must be an isomorphism. It is also know that $H_q(A)$ is a normal subgroup of $H_q(X)$ and $H_q(X,A)$ is isomorphic to $H_q(X)/H_q(A)$. My idea is consider that mapping: $$\begin{array}{rcll} \Phi:&H_q(X)&\longrightarrow &H_q(A)\oplus H_q(X,A)\\ &[T]&\longmapsto &\Phi[T]=H_q(r)[T]+pr[T] \end{array}$$ and prove that $\Phi$ is an isomorphism. Is it a good idea?

(we denote $pr$ as the projection $H_q(X)\to H_q(X)/H_q(A)$).

Note: This question has been solved here, but I can't understand anything.

  • 0
    My idea is to show that $r_*:H_n(X,A)\rightarrow H_n(A,A)$ is an isomorphism. – 2017-01-14
  • 0
    Note that $H_n(A,A)=0$. – 2017-01-14
  • 0
    If we show that $H_q(X,A)\cong H_q(A,A)$, what we can conclude? – 2017-01-14
  • 0
    Use the pair sequence. – 2017-01-15

1 Answers 1

1

It is probably a cleaner idea to use some of these exact sequences and homological algebra you have floating around.

What is the definition of a retraction? It is $r:X\rightarrow A$ such that $r\circ i=id_A$. If we look at the maps $i_\ast$ and $r_\ast$ induced by our inclusion and retraction on singular homology we obtain the equation, $$r_\ast\circ i_\ast=(r\circ i)_\ast=id_\ast=id,$$ which implies that $i_\ast$ is injective. From the long exact sequence of a pair (http://mathworld.wolfram.com/LongExactSequenceofaPairAxiom.html) we see that the fact that $i_\ast$ is injective, implies that $\partial=0$ is the zero map. This gives us a short exact sequence, $$0\longrightarrow H_k(A)\longrightarrow H_k(X)\longrightarrow H_k(X,A)\longrightarrow 0.$$ If you look on page 147 of Hatcher (https://www.math.cornell.edu/~hatcher/AT/AT.pdf) and the splitting lemma there, you see that we are in the situation of part (a). Hence our short exact sequence splits.

This is an easy argument if you can use these fundamental tools of homology, otherwise things tend to get a little messy.