One point compactification. (Pushout) 
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 $(V\times W)^{\infty}$ 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 $A\subset B$ a closed subspace with inclusion function $\alpha:A\rightarrow B$, and continuous mapping $f:A\rightarrow X$. 
We define the equivalence relation "$\sim$" on $X\coprod B$ as $inc_X\circ f(a)\sim inc_B\circ\alpha(a)$ for each $a\in A$. We thus define $Y:=(X\coprod Y)/\sim$ (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 $g:Y\rightarrow W$, 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 $(V\times W)^{\infty}$
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)
 A: Have a look at the catgorical definition of pushout. It is enough to show that $(V\times W)^\infty$ has the universal property. That's because pushouts are unique up to isomorphism.
I will use $V^\infty\vee W^\infty$, i.e. the wedge sum instead of $(V^\infty\times\{\infty\})\cup(\{\infty\}\times W^\infty)$ because thats what it is.
So we have a commutative diagram
$$\begin{CD}
V^{\infty}\vee W^{\infty} @>{i}>> V^{\infty}\times W^{\infty}\\
@VVV @VV{f}V\\
\{\infty\} @>{g}>> (V\times W)^\infty
\end{CD} $$
So $i$ is the inclusion, $g(\infty)=\infty$ and
$$f(v,w)=\left\{\begin{matrix}\infty &\text{if }v=\infty\text{ or }w=\infty \\ (v,w) &\text{otherwise}\end{matrix}\right.$$
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
$$\phi:(V\times W)^\infty\to P$$
$$\phi(v,w)=a(v,w)$$
$$\phi(\infty)=b(\infty)$$
So obviously $\phi\circ f=a$ and $\phi\circ g=b$. 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.
