Wedge sum of spheres is the quotient $X^n/X^{n-1}$

Question

As in the title, I want to prove that $\bigvee_jS_j^n=X^n/X^{n-1};\ X$ is a $CW$ complex and $X^n$ and $X^{n-1}$ are the $n-$ and $n-1$-skeleta. Below, I present a sketch of an attempt using pushouts, but it is not correct. I would like to know if my idea can be made to work, and I would also like to see a proof that does not use category theory.

The wedge sum can be viewed as a adjunction space, via the map $f:\{x_j\}\mapsto \{*\}$ and then $\bigvee_jS_j^n$ is a pushout:

$\tag 1\require{AMScd} \begin{CD} \{x_j\} @>>> \{*\}\\ @V{i}VV @VV{q_1}V\\ \coprod_j S^n @>>{q_2}> \bigvee_jS_j^n\end{CD}$

where $q_{1,2}=p\circ i_{1,2}$ and $p$ is the canonical quotient map.

We also have the pushout

$\tag2 \require{AMScd} \begin{CD} \coprod_jS^{n-1}_j @>{\Phi}>> X^{n-1}\\ @V{i}VV @VV{q_1}V\\ \coprod_jD^{n}_j @>>{q_2}> X^n\end{CD}$

$q_{1,2}=p\circ i_{1,2}$ and $p$ is the canonical quotient map and $\Phi$ is the characterstic map.

so if we adjoin to this square, the square

$\tag 3\require{AMScd} \begin{CD} X^{n-1} @>>> \{*\}\\ @V{i}VV @VV{q_1}V\\ X^{n} @>>{q_2}> X^n/X^{n-1}\end{CD}$,

we get another pushout

$\tag4\require{AMScd} \begin{CD} \coprod_jS^{n-1}_j @>>> \{*\}\\ @V{i}VV @VVV\\ \coprod_jD^{n}_j @>>> X^n/X^{n-1}\end{CD}$.

I want to compare $(1)$ and $(4)$ to get the isomporphism, but of course, it does not work.

Answer 1

I think you only need to write another pushout square saying that the $n$-sphere is the quotient of the $n$-ball by the $(n-1)$-sphere :

$$\begin{CD} S^{n-1} @>>> \{*\}\\ @VVV @VVV\\ D^n @>>> S^n.\end{CD}$$

The disjoint union of such pushout squares will give you :

$$\begin{CD} \coprod_{j \in J}S^{n-1}_j @>>> J\\ @V{i}VV @VVV\\ \coprod_{j \in J}D^{n}_j @>>> \coprod_{j \in J}S^{n}_j\end{CD}$$

This, glued to your first square, gives exactly : $$\begin{CD} \coprod_{j \in J}S^{n-1}_j @>>> \{*\}\\ @V{i}VV @VVV\\ \coprod_{j \in J}D^{n}_j @>>> \bigvee_jS_j^n\end{CD}$$ Compare this to (4) to get the result you were seeking.

Answer 2

Alternate answer : $X_n/X_{n-1}$ is a CW-complex (quotient of a CW-complex by a subcomplex) with one 0-cell and only n-cells, and this must be a wedge of n-spheres.

Answer 3

Note the standard diagram: $$\require{AMScd}\begin{CD}S^{n-1}@>>>\{*\}\\@VVV@VVV\\D^n@>>>S^n\end{CD}$$

Taking disjoint unions, we see that $$\begin{CD}\coprod S^{n-1}_j@>>>\{x_j\}\\@VVV@VVV\\\coprod D^n_j@>>>\coprod S^n_j\end{CD}$$ is a pushout. Combining this with (1), we get $$\begin{CD}\coprod S^{n-1}_j@>>>\{*\}\\@VVV@VVV\\\coprod D^n_j@>>>\bigvee S^n_j\end{CD}$$ is a pushout. Combining with (4), we get our desired conclusion.

A categoryless proof is the following:

Observe that $X^n\setminus X^{n-1}$ is a disjoint union of open disks. The boundary of each disk is contained in $X^{n-1}$, so we have an obvious map from $X^n$ to the wedge sum of spheres by sending each open disk in the complement of $X^{n-1}$ homeomorphically to the complement of the wedge point in one of the spheres and $X^{n-1}$ to the wedge point. This is clearly surjective. Moreover, if $U$ is open in the wedge sum, then if it doesn't contain the wedge point, its preimage is a disjoint union of open sets each contained in some open disk of $X^n\setminus X^{n-1}$, and hence is open. Otherwise if it contains the wedge point, its preimage contains $X^{n-1}$. However to be open in the wedge sum, that means that the intersection of $U$ with every sphere is open, so its preimage in $X^n$ has open intersection with every closed $n$-disk. Thus by the definition of the topology on $X^n$, the preimage of $U$ is open in this case as well. Thus our map is continuous. The map is also closed, since if $K$ is closed in $X^n$, then $K\cap D^n_j$ is closed for each $j$, so the image of $K$ in each sphere being wedged together is closed, since the standard quotient $D^n\to S^n$ is a closed map (since $D^n$ is compact). Thus the image of $K$ is closed in the wedge of spheres. Thus the map is a quotient map, giving the desired homeomorphism. $X^n/X^{n-1} \cong \bigvee S^n$.