Jtdylktuy

I apologize in advance for the formatting. I am new to commutative diagrams in latex.

Let $V$ and $W$ be finite dimensional vector spaces over $mathbb{R}$, with the pushout

$require{AMScd}$
begin{CD}
(V^{infty}times {infty})cup({infty}times W^{infty}) @>{}>> V^{infty}times W^{infty}\
@VVV @VVV\
{infty} @>{}>> Y
end{CD}

Show that $(Vtimes W)^{infty}$ is homeomorphic to Y

Here $"infty"$ denotes the 1-point compactification of a set

Previously, we proved a few remarks about pushouts.

If $X$ and $B$ are topological spaces and $Asubset B$ a closed subspace with inclusion function $alpha:Arightarrow B$, and continuous mapping $f:Arightarrow X$.

We define the equivalence relation "$sim$" on $Xcoprod B$ as $inc_Xcirc f(a)sim inc_Bcircalpha(a)$ for each $ain A$. We thus define $Y:=(Xcoprod Y)/sim$ (quotient space).

It was shown (universal property of the pushout) that given a commuting diagram marked in the below picture of solid maps/arrows, there exists a unique mapping $g:Yrightarrow W$, which makes the whole diagram commutative.

I want to somehow use this property (I think) to show there exists a homoemorphism $h$ from $Y$ to $(Vtimes W)^{infty}$

I honestly have no clue how to approach this problem, and would appreciate a hint or two to guide me in the right direction. Any help would be much appreciated!

Universal property of the pushout (picture)

Have a look at the catgorical definition of pushout. It is enough to show that $(Vtimes W)^infty$ has the universal property. That's because pushouts are unique up to isomorphism.

I will use $V^inftyvee W^infty$, i.e. the wedge sum instead of $(V^inftytimes{infty})cup({infty}times W^infty)$ because thats what it is.

So we have a commutative diagram

$$begin{CD}
V^{infty}vee W^{infty} @>{i}>> V^{infty}times W^{infty}\
@VVV @VV{f}V\
{infty} @>{g}>> (Vtimes W)^infty
end{CD} $$

So $i$ is the inclusion, $g(infty)=infty$ and

$$f(v,w)=left{begin{matrix}infty &text{if }v=inftytext{ or }w=infty \ (v,w) &text{otherwise}end{matrix}right.$$

So this is the first part of the universal property. I encourage you to check that the diagram commutes and that $f$ is continuous.

Lets have a look at the second part of the universal property. Assume that
$$begin{CD}
V^{infty}vee W^{infty} @>{i}>> V^{infty}times W^{infty}\
@VVV @VV{a}V\
{infty} @>{b}>> P
end{CD} $$

is some commutative diagram. Define

$$phi:(Vtimes W)^inftyto P$$
$$phi(v,w)=a(v,w)$$
$$phi(infty)=b(infty)$$

So obviously $phicirc f=a$ and $phicirc g=b$. So all that is left is to show that $phi$ is continuous. Can you complete the proof?

Side note: If you look carefuly at each function I've defined you will see that there's no magic here. These are just natural choices.