Realization of Simplicial Sets respects Product My question refers to an argument in the the proof of thm 11.6 in Laures' and  Szymik's "Grundkurs Topologie" (page 227). Sorry, but there exist only a German version. Here the relevant excerpt:

The problem occurs in Step 2 ("Schritt 2"):
We use following notations: 
If $K$ is a simplicial set then $\vert K \vert := \coprod_{n \ge 0}K_n \times \Delta^n_{top}$ is the realization of $K$, $\Delta^n_{top}$ is the standard $n$-simplex and $\Delta^n= Hom_{\Delta}(-, [n])$.
The aim (in "Schritt 2")is to show that for all simplicial sets $X$ (= functors from $\Delta$ to $(Set)$) the canonical map 
(*)$$\vert X \times \Delta^n \vert = \vert X \vert \times \vert \Delta^n \vert$$ is homeomorphism. 
So firsty we fix $X$ and observe that the class $C$ of all $X$ for which (*) is a homeomorphism is closed under pushouts and sums $\coprod$.
Then we consider the category $S(X)$ of all simplicial maps $\Delta^n \to X$. 
The morphism of this category are the commutative triangles as given at page 227.
In the next step the author claims that following diagram is a pushout:

Why? I don't see why it has the universal property of a pushout. 
 A: Let $Y$ be any simplicial set and assume we have two maps $f,g : \displaystyle\coprod_{x: \Delta^n \to X}\Delta^n \to Y$ that make the above diagram commute, that is $f\circ (id, j) = g\circ (id, i)$ .
I will use the following convention : an element of the coproduct $\coprod_{i\in I}X_i$ is denoted $(x,i)$, where $x\in X_i$
Now assume $h: X\to Y$ makes the whole thing commute. Let $x\in X_n$ be an $n$-simplex. Then $x$ corresponds to some $\tilde{x} : \Delta^n \to X$ (by the Yoneda lemma), and so $h_n(x) = f((id_{[n]},\tilde{x}))(=g((id_{[n]},\tilde{x}))$ by commutation of the diagram). Therefore if $h$ exists it is unique. 
Now define $h$ degree-wise as above, with $f$ : it is well-defined on each degree by the Yoneda lemma. We must show that it is a simplicial map and that it makes the diagram commute, after this by the uniqueness above we will be done. That it makes the diagram commute is quite obvious, as $(id,i),(id,j)$ are surjective, so since $f\circ (id,j) = g\circ (id,i)$ and $h$ is defined through $f$, so this is clear. 
Let's now prove that it is a simplicial map. Let $\varphi : [m]\to [n]$ be nondecreasing and $x\in X_n$. Let me write the action of $\varphi$ on the right, as $X$ is a contravariant functor of $[k]$. 
Then $h(x\cdot \varphi) = f((id_{[n]},\widetilde{x\cdot\varphi})$, but $\widetilde{x\cdot \varphi}$ is nothing but $\tilde{x}\circ \overline{\varphi}$ where $\overline{\varphi}$ is the induced map $\Delta^m\to \Delta^n$; for this you have to explicit the Yoneda isomorphism.
Also, $(id_{[n]},\tilde{x}\circ \overline{\varphi}) = (id,j)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$ where $y = x\cdot \varphi$, therefore $h(x\cdot \varphi) = g((id,i)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$. 
Now $i(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x}) = (\overline\varphi(id_{[m]}), \tilde{x})$ by definition. 
Now if you recall the definition of the induced map $\Delta^m\to \Delta^n$ and of the simplicial structure on $\Delta^n$, you see that $\overline\varphi (id_{[m]}) = id_{[n]}\cdot \varphi$. Therefore $h(x\cdot \varphi) =g((id_{[n]}, \tilde{x})\cdot \varphi) = g((id_{[n]},\tilde{x}))\cdot \varphi = h(x)\cdot \varphi$ : $h$ is simplicial; and we are done. 
