# Telescope of CW-complexes

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?

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 $$\bigcup_{n \text{ odd}} \bar X_n \approx \bigvee_{n \text{ odd}} X_n$$.

• Thank you for your answer. I have a couple of (maybe trivial) questions though: why does $X’_{n}$ deformation retracts to $\bar{X_n}$? And why is this last one an homeomorphic copy of $X_n$?
– cip
Commented Feb 11, 2021 at 19:22
• $\bar X_n$ is the image of the embedding $X_n \to X', x \mapsto [n,x]$, hence it is a homeomorphic copy of $X_n$. And the "projection" $p_n : [n-1,n] \times X_n \to \{n\} \times X_n, p_n(t,x) = (n,x)$, is a strong deformation retraction via $R_n : [n-1,n] \times X_n \times I \to [n-1,n] \times X_n, R_n(t,x,s) = (s(n-t) + t ,x)$. It maps $[n-1,n] \times * \times I$ to $(n,*)$ and thus induces a pointed strong deformation retraction $X'_n \to \bar X_n$. Commented Feb 11, 2021 at 23:43