# Constructing the geometric realization of a simplicial set

I'm trying to understand the construction of the geometric realization of a simplicial set. The goal is to solve the following problem:

For each simplicial set $$S_\bullet$$, find a space $$|S_\bullet|$$, called a geometric realization of $$S_\bullet$$, and a natural isomorphism $$\hom(S_\bullet, \mathrm{Sing_\bullet}(-))\cong \hom(|S_\bullet|, -)$$.

Here, $$\mathrm{Sing}_\bullet$$ is the singular simplicial set functor.

I have several questions:

1. The solution can be summarized as follows: firstly, show that whenever a class of simplicial sets has a geometric realization, then each simplicial set that can be built as a colimit of these simplicial sets has a geometric realization too. Secondly, the $$\Delta^n$$s have a geometric realization by the Yoneda lemma. Thirdly, by the co-Yoneda lemma every simplicial set is the colimit of $$\Delta^n$$s. I have a hard time making the first step precise. The obvious way of writing it down is as follows:

Let $$F\colon J\to \mathbf{sSet}$$ and suppose each $$F(j)$$ has a geometric realization $$|F(j)|$$. Then $$\mathrm{colim}_j |F(j)|$$ is a geometric realization of $$\mathrm{colim}_j F(j)$$.

But what is $$\mathrm{colim}_j |F(j)|$$? It should be the colimit of some functor with domain $$J$$. But what functor? I'm tempted to say $$|-|\circ F\colon J\to \mathbf{Top}$$ -- but we are only in the process of constructing $$|-|$$!

1. Suppose the above problem is solved, i.e., we have constructed $$|S_\bullet|$$ for each simplicial set $$S_\bullet$$. And in each case we have constructed a natural bijection $$\alpha_{S_\bullet}\colon \hom(S_\bullet, \mathrm{Sing_\bullet}(-))\cong \hom(|S_\bullet|, -).$$ How can we now built a left adjoint of $$\mathrm{Sing}_\bullet$$? (We already did it on objects and now we do it on morphisms.)

I heard that there is always a unique way to define the morphisms of the left adjoint. But I'm confused, for the following reason: let $$S_\bullet \to T_\bullet$$ be a map of simplicial set. We want to construct a morphism $$|S_\bullet|\to |T_\bullet|$$. By Yoneda, it suffices to construct a mapping $$\hom(|T_\bullet|, -)\to \hom(|S_\bullet|, -).$$ Here is how we do it: for each $$X$$, given $$|T_\bullet|\to X$$, use $$\alpha_{T_\bullet}^{-1}$$ to obtain a morphism $$T_\bullet\to \mathrm{Sing}_\bullet(X)$$. Now precompose with $$S_\bullet\to T_\bullet$$ to get a morphism $$S_\bullet \to \mathrm{Sing}_\bullet(X)$$. Finally, apply $$\alpha_{S_\bullet}$$ to get a morphism $$|S_\bullet|\to X$$.

But is this way of defining the functor $$|-|$$ on morphisms really unique? After all, it might depend on $$\alpha_{S_\bullet}$$!

1. Is there a more elegant way to describe $$|-|$$ on morphisms? This reminds me of a quote I read in Riehl's A Leisurely Introduction to Simplicial Sets:

Uniqueness of the universal property will imply that $$L$$ is functorial, as is always the case when one uses a colimit construction to define a functor.

I have no idea what this means, but I'm very curious. Can you formulate the general principle behind that quote precisely?

1. You've basically got it. $$|-|\circ F$$ is perfectly well defined since you assumed $$F$$ is valued in simplicial sets for which $$|-|$$ is defined. However, note that you need $$|-|\circ F$$ to also be defined on morphisms in $$J.$$ This ties into...
2. No, it is not really unique. It depends on a choice of $$\alpha_{S_\bullet},$$ but only on that. The best way to proceed is to strengthen your construction in 1: assume that "a simplicial set has a geometric realization" means you've defined $$|S_\bullet|$$ along with a choice of $$\alpha_{S_\bullet}.$$ With that clarification, your concerns in both 1 and 2 are resolved.
3. Given $$f:S_\bullet\to T_\bullet$$ and $$x\in S_\bullet,$$ let $$x$$ be represented by $$a_x:\Delta^n\to S_\bullet.$$ Then we have $$|f|\circ |a_x|=|f\circ a_x|.$$ Since $$f\circ a_x:\Delta^n\to T_\bullet$$ represents the simplex $$f(x),$$ we see that $$|f|$$ sends the realization of $$x,$$ which is some quotient space of $$|\Delta^n|$$ in $$|S_\bullet|,$$ to the realization of $$f(x).$$ In other words, the geometric realization of a simplicial set map acts on simplices, now geometrically visible, in the same way as the original map did.
I think what Riehl's quote refers to is that, in general, if you have a functor $$F:A\to\mathcal C$$ with $$A$$ small and $$\mathcal C$$ cocomplete, you always get a cocontinuous functor $$\widehat F:\widehat A\to \mathcal C$$ via the co-Yoneda lemma. This is closely related to what we're doing here, except that we focused on defining geometric realization using the intended right adjoint. You can get away without this by starting with the geometric realization of maps between representable simplicial sets, which resolves the missing morphisms from (1) in a different manner.
1. You are right. If one wants to be really precise, one needs to assume that $$|-|$$ has already been defined on some subcategory of $$\operatorname{Set}_{\Delta}$$. Call this subcategory $$\mathcal C$$. Then your formula makes sense provided that the diagram $$F\colon J\to \operatorname{Set}_{\Delta}$$ factors through $$\mathcal C$$, which is how the task is to be understood. And indeed, one only really needs the case where $$\mathcal C$$ is $$\Delta\hookrightarrow\operatorname{Set}_{\Delta}$$, where the functor $$|-|\colon\Delta\to\operatorname{Top}$$ is the evident one.
2. and 3. The trick is to not try to define $$|-|$$ directly, but to use Yoneda's lemma: the functor $$\hom(-,\operatorname{Sing}_\bullet(-))\colon \operatorname{Set}_{\Delta}^{\operatorname{op}}\times\operatorname{Top}\to\operatorname{Set}$$ can be equivalently regarded as a functor $$\operatorname{Set}_{\Delta}^{\operatorname{op}}\to \operatorname{Fun}(\operatorname{Top},\operatorname{Set})$$ that carries a simplicial set $$S_\bullet$$ to $$\hom(S_\bullet,\operatorname{Sing}_\bullet (-))\simeq \hom(|S_\bullet|,-)$$. Therefore, the image of $$S_\bullet$$ is a representable functor, which implies that this functor factors through the Yoneda embedding $$\operatorname{Top}^{\operatorname{op}}\hookrightarrow \operatorname{Fun}(\operatorname{Top},\operatorname{Set})$$. In this way, one obtains a functor $$|-|\colon \operatorname{Set}_{\Delta}^{\operatorname{op}}\to\operatorname{Top}^{\operatorname{op}}$$, which upon taking opposite categories yields the geometric realisation functor.