Let $Z=$pushout$(Y\leftarrow A\rightarrow X)$, then prove that $Z\times W=$pushout$(Y\times W\leftarrow A\times W\rightarrow X\times W)$

Question

Let $Z$ be the pushout of the diagram

$\require{AMScd}$ \begin{CD} A @>{f}>>X\\ @V{g}VV @VVV\\ Y @>>> Z \end{CD}

$Z$ is defined to be $Z:=(X\sqcup Y)/f(a)\sim g(a)\ \forall a\in A$

Let $W$ be any space. Then Prove that $Z\times W$ is pushout of the following diagram

$\require{AMScd}$ \begin{CD} A\times W @>{f\times id}>>X\times W\\ @V{g\times W}VV @VVV\\ Y\times W @>>> Z\times W \end{CD}

i.e. we have to prove that $Z\times W\approx (X\times W\sqcup Y\times W)/(f(a),w)\sim (g(a),w)\ \forall a\in A, w\in W$

Let $i:X\to Z$ denotes the usual map $x\mapsto [x]$ and $j:Y\to Z$ denotes the map $y\mapsto [y]$. Then we have maps $i_1=i\times id:X\times W\to Z\times W$ and $j_1=j\times id:Y\times W\to Z\times W$. It's easy to see as $i,j$ are injective $i_1,j_1$ are injective.

Then from the universal property of disjoint union we have a continuous map $\phi:(X\times W)\sqcup(Y\times W)\to Z\times W$, this map is onto as well. As $i_1,j_1$ are injective and $\phi(f(a),w))=\phi(g(a),w)$. From the universal property of quotient topology, we have a continuous bijective map $$\tilde{\phi}:(X\times W)\sqcup (Y\times W)/\sim\to Z\times W$$

To show $\tilde{\phi}$ is a homeomorphism, we have to prove that $\phi$ is a quotient map. But I cannot prove that.

Can anyone help me to finish the proof? Thanks for the help in advance.

Edit: If I take the construct the inverse of $\tilde{\phi}$ like this-

$\psi:Z\times W\to Z'$ by $\psi(([x],w))=[(x,w)]$ and $\psi(([y],w))=[(y,w)]$ (Here $Z'=(X\times W\sqcup Y\times W)/\sim$

But how to show $\psi $ is continuous?