If $X_{\lambda}$ is a family of CW complexes, then $\tilde H_*(\lor_{\lambda}X_{\lambda}) \approx \sum \tilde H_*(X_{\lambda})$

Question

If $\{X_{\lambda}: \lambda \in \Gamma\}$ is a family of CW complexes with basepoint, then $\tilde H_*(\lor_{\lambda}X_{\lambda}) \approx \sum \tilde H_*(X_{\lambda})$

$(\lor_{\lambda}$ means wedge)

I'm having trouble figuring a way to show this.

I can see I'd need to find an isomorphism that maps: $$u^n + B_n(\lor_{\lambda}X_{\lambda}) \mapsto \sum_{\lambda}(\sigma_{\lambda}^n + B_n(X_{\lambda}))$$where $u^n : \Delta^n \rightarrow \lor_{\lambda}X_{\lambda}$ and $\sigma_{\lambda}^n : \Delta^n \rightarrow X_{\lambda}$, but I can't figure a way to connect these.

Anyone have any ideas?

Answer 1

It follows from the following:

Proposition. Let $X$ be a topological space and $A$ a closed subspace which is a deformation retract of some open neighbourhood. Then the quotient map $\phi:X\to X/A$ induces an isomorphism $$\phi_{n}:H_n(X,A)\to H_n(X/A, A/A)\simeq \tilde H_n(X/A)$$ for all $n$.

which is a consequence of the excision theorem. You can find the proof in Allen Hatcher's "Algebraic Topology", Proposition 2.22.

Now the statement follows by realizing that if $X$ is a CW-complex then every point is a closed, neighbourhood deformation retract in $X$. Now consider the canonical quotient map:

$$q:\bigsqcup X_\lambda\to\bigvee X_\lambda$$

with a disjoint union on the left side. The proposition applies which yields an isomorphism

$$H_n(\bigsqcup X_\lambda, A)\simeq\tilde H_n(\bigvee X_\lambda)$$

where $A$ is the set of all basepoints. It seems that the left side can be further reduced to

$$H_n(\bigsqcup X_\lambda, A)\simeq \bigoplus H_n(X_\lambda, \{*\})\simeq \bigoplus \tilde H_n(X_\lambda)$$

The second isomorphism is obvious. The first one is a relative version of "homology of a disjoint union is a direct sum of homologies". I guess you should look for details in Hatcher's book.