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.
Best Answer
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.