Q:

Simplicial set as a colimit

Let $K \in \mathrm{Set}_{\Delta}$ be a simplicial set. Then $K$ is the colimit of the diagram 
  $$F \colon \Delta/K \to \mathrm{Set}_{\Delta} \, $$
  that assigns to each $\Delta^n \to K$ the standard $n$-simplex $\Delta^n$. 

This should be just about unravelling the definition of a limit cone, but I don't know how to rigorously prove this. A cone under $F$ is an object $x \in \mathrm{Set}_{\Delta}$ together with a natural transformation $\eta \colon F \Rightarrow \Delta( x)$, where $\Delta(x)$ stands for the diagonal functor.
$K$ is a cone under $F$, since a natural transformation $\eta \colon F \Rightarrow \Delta(K)$ is just a series of morphisms $\eta_f \colon F(\Delta^n \xrightarrow{f} K) \to K$, that is, a map $\Delta^n \to K$, which is given by $\eta_f = f$ (though I'm probably wrong). But how is that cone under $F$ a colimit cone? 
Thank you in advance.  

A:

Your definition of $\eta : F \Rightarrow \Delta(K)$ is correct (though you should check that it is natural—this is an easy task!)
Let $X$ be a simplicial set and let $\theta : F \Rightarrow \Delta(X)$ be another cone. Then you can define $u : K \to X$ as follows. First note that there are natural isomorphisms $K_n \cong \mathrm{Set}_{\Delta}(\Delta^n, K)$ and $X_n \cong \mathrm{Set}_{\Delta}(\Delta^n, X)$ by the Yoneda lemma; and then note that there is a map $\mathrm{Set}_{\Delta}(\Delta^n, K) \to \mathrm{Set}_{\Delta}(\Delta^n, X)$ given by $f \mapsto \theta_f$. This induces functions $u_n : K_n \to X_n$ for each $n \in \mathbb{N}$, which piece together to form a map of simplicial sets $u : K \to X$.
Finally, check that $\Delta(u) : \Delta(K) \Rightarrow \Delta(X)$ is the unique natural transformation such that $\Delta(u) \circ \eta = \theta$.

