Wedge sum of spheres

Question

Let's $X$ be a CW-complex. If $X^{(n)}$ is the n-skeleton of $X$ and $\Lambda_n$ is a set of index. How could I prove that $X^{(n)}/X^{(n-1)}=\bigvee_{\alpha \in \Lambda_n} S^n_{\alpha}$? Thank you for your help!

Answer 1

The $n$-skeleton $X^n$ is obtained from the $(n-1)$-skeleton by gluing a union of $n$-balls along their boundaries to that $X^{n-1}$. This gives the first of the two pushout squares in the diagram. $\require{AMScd}$ $$\begin{CD} \coprod\nolimits_\Lambda S^{n-1}_\alpha @>\varphi>> X^{n-1} @>>> \{*\} \\ @VVV @VVV @VVV\\ \coprod\nolimits_\Lambda D^n_\alpha @>>> X^n @>>> X^n/X^{n-1} \end{CD}$$ Since the composition is a pushout, this shows that $X^n/X^{n-1}$ is obtained by gluing the $n$-balls along their boundaries to a single point.

Answer 2

Use Yoneda lemma in the pointed category $Top_*$ and a bit of fancy: $Top_*(X^{(n)}/X^{(n-1)}, Y) \simeq \{ f:X^{(n)} \to Y \ s.t. f_{| X^{(n-1)}} = constant \} \simeq \{f_{\alpha}: D^n \to Y \ s.t. f_{\alpha| S^{n-1}} = * \}_{\alpha \in \Lambda _n} \simeq Top_*(S^n,Y)^{\Lambda _n} \simeq Top_*(\bigvee _{\alpha \in \Lambda _n} S^n, Y) $ .