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?

  • $\begingroup$ 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

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.


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.


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