5
$\begingroup$

This is an exercise of Bredon (pg. 190) which I tried to do but got stuck at one part. He asks the following:

Let $X$ be a Hausdorff space and let $x_0 \in X$ be a point having a closed neighbourhood $N$ in $X$, of which $\{x_0\}$ is a strong deformation retract. Let $Y$ be a Hausdorff space and let $y_0 \in Y$. Define $X \vee Y=X \times \{y_0\} \cup \{x_0\} \times Y$. Show that the inclusion maps induce isomorphisms $\widetilde{H}_i(X)\oplus \widetilde{H_i}(Y) \rightarrow\widetilde{H}_i(X \vee Y)$ for any homology theory, whose inverse is induced by the projections of $X \vee Y$ to $X$ and $Y$.

OBS: Note that this is not a duplicate of this question, since we only assume one of the spaces form a "good pair" with the point.

I proceeded as follows:

The inclusions $i:(X\times \{y_0\},\{x_0 \times y_0\}) \rightarrow (X \vee Y,\{x_0 \times y_0\})$ and $j: (X \vee Y, \{x_0 \times y_0\}) \rightarrow (X \vee Y, X \times \{y_0\})$ induce a long exact sequence in the homology of the triple $(\{x_0 \times y_0\},X\times \{y_0\} ,X \vee Y )$:

$\require{AMScd}$ \begin{CD} \cdots H_*(X\times \{y_0\},\{x_0 \times y_0\}) @>i_*>> H_*(X \vee Y,\{x_0 \times y_0\}) @>j_*>> H_*(X \vee Y, X \times \{y_0\}) \cdots \\ \end{CD}

The conditions over $X$ allow us to conclude (by excision and homotopy) that $H_*(X \vee Y, X \times \{y_0\}) \cong H_*(\{x_0\} \times Y, \{x_0 \times y_0\})$ -- (1). If I show that this is actually a short exact sequence which is split, then the result follows.

Since $\pi_1 \circ i=Id$, then $\pi_1^* \circ i^*$ is an isomorphism, and it follows that $i$ is injective (functorial property).

What I would wish now is to use see the right side as induced by the projection $\pi_2$ (when seen under the isomorphism (1)). I think that everything would follow by that. But the isomorphism in (1) does not make it clear to me that this is the case. This is where I got stuck. Any hints?

$\endgroup$
  • $\begingroup$ I have your same problem, but I don't understand where it's used the fact that $N$ is S.D.R of $x_0$. Can you explain me? $\endgroup$ – Vincenzo Zaccaro Dec 28 '16 at 18:04
5
$\begingroup$

An easier approach would be to use the reduced Mayer-Vietoris sequence (which exists in arbitrary homology theories) as follows:

We can write $X\vee Y$ as a union of the two open subsets $U=X\cup N$ and $V=Y\cup N$. Note that $U$, respectively $V$, deformation retract onto X, respectively $Y$. Moreover, the intersection $U\cap V$ deformation retracts onto a point.

Thus, the Mayer-Vietoris sequence corresponding to the open cover $X=U\cup V$ reads:

$$\cdots \to \underbrace{\widetilde H_i(\text{pt})}_{=0}\to \widetilde H_i(X)\oplus \widetilde H_i(Y)\to \widetilde H_i(X\vee Y)\to \underbrace{ \widetilde H_{i-1}(\text{pt})}_{=0}\to\cdots$$ and the middle map is induced by the inclusions $X,Y\hookrightarrow X\vee Y$. This provides the desired isomorphism.

$\endgroup$
0
$\begingroup$

You nearly have solved your problem. Observe the connecting homomorphisms $\partial_*$ of the long exact sequence you used $$\require{AMScd} \begin{CD} \cdots @>{\partial_*}>> H_i(X,pt) @>{i_{1*}}>> H_i(X \vee Y,pt) @>{}>> H_i(X \vee Y, X) @>{\partial_*}>> \cdots\\ @.@.@VV\pi_{2*}V @A{\cong}Ak_*A \\ @.@.H_i(Y,pt)@= H_i(Y, pt) \end{CD} $$ Since you can see $i_{1*}: H_i(X,pt)\to H_i(X \vee Y,pt)$ is injective, we just need to observe that by the exactness $\text{Im} ~\partial_* (=\text{Ker}~ i_{1*})$ are all trivial. Then, combining your argument(i.e.isomorphism(1)) we can get a short exact sequence as you desired: $$\require{AMScd} \begin{CD} 0 @>{}>> H_i(X,pt) @>{i_{1*}}>> H_i(X \vee Y,pt) @>{\pi_{2*}}>> H_i(Y,pt) @>{}>> 0 \end{CD} $$

What is more, our argument can further answer the second part of the problem as follows. Since $\pi_1∘i_1=Id$ and thus $\pi_{1*}∘i_{1*}=Id$, the short exact sequence that we just get is splitting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.