Telescope of CW-complexes

Question

Let $X_n$ the n-th term of an ascending sequence of pointed CW-complexes in their homotopy category. Let $X’= \bigcup_{n\geq 1} [n-1, n]^{+} \wedge X_n$ and let $A_1 = \bigcup_{k\geq 1, k odd} [k-1, k]^{+} \wedge X_k$ and $A_2 = \bigcup_{k\geq 0, k even} [k-1, k]^{+} \wedge X_k$.

Why is it true that $A_1 \cap A_2 = \bigvee_k X_k$ and $A_1$ is homotopically equivalent to $\bigvee_{k,odd} X_k$?

This is my very first approach to smash products and wedge sums, so I might be failing to understand some fundamental aspects of their definition. Any hints?

Answer 1

We have an ascending sequence of pointed CW-complexes $X _1\subset X_2 \subset X_3 \subset \ldots$ with a common basepoint $* \in X_1$. Noting that $X'_n = [n-1,n]^+ \wedge X_n = ([n-1,n] \times X_n) / ([n-1,n] \times *)$, we see that $X'_n$ deformation retracts to the subspace $\bar X_n = \{n\} \times X_n' \subset X'$ which is a homeomorphic copy of $X_n$.

1. The subspace $\bigcup_n \bar X_n \subset X'$ is nothing else than the disjoint sum of the $X_n$ with all basepoints $* \in X_n$ identified to a single point, i.e. $\bigcup_n \bar X_n \approx \bigvee_n X_n$.

2. Similarly the subspace $A_ 1 = \bigcup_{n \text{ odd}} X'_n \subset X'$ is nothing else than the disjoint sum of the $X'_n$, $n$ odd, with all basepoints $* \in X'_n$ identified to a single point, i.e. $\bigcup_{n \text{ odd}} X'_n \approx \bigvee_{n \text{ odd}} X'_n$.

3. By construction $A_1 \cap A_2 = \bigcup_n \bar X_n$, i.e. $A_1 \cap A_2 \approx \bigvee_n X_n$.

4. Since each $X'_n$ deformation retracts to $\bar X_n$, we see that $A_1$ deformation retracts to $\bigcup_{n \text{ odd}} \bar X_n \approx \bigvee_{n \text{ odd}} X_n$.