Colimit of simplicial set in bijection with path connected components of geometric realization

Question

I want to show that if $X$ is a simplicial set we have a bijection $\text{colim}_{\Delta^{op}}X\cong \pi_0|X|$ with $|X|$ the geometric realization of $X$. So here $|X|=\coprod_{n\geq 0}X_n\times \Delta^n/\sim$ where $\Delta^n$ is the standard $n$-simplex and $\sim$ generated by $(x,\delta^{i}(t))\sim(d_i(x),t)$, $(x,\sigma^{i})\sim(s_i(x),t)$.

Here I have proved what if $X$ is a simplicial set, $\text{colim}_{\Delta^{op}}X\cong\text{coeq}(X_1\rightrightarrows X_0)$ and we have a way of seeing it as $X_0/ \sim $ where $x\sim x'$ if $\exists y\in X_1$ such that $d_0(y)=x,d_1(y)=x'$. So we can find just find a bijection between $X_0/\sim$ and $\pi_0(|X|)$.\

I thought about sending $[x]$ to $[(x,1)]$ with $(x,1)\in X_0\times \Delta^0$ but I'm not sure it works.

Do you have ideas ?

Answer 1

You seem to unfold the definitions correctly and be on track to a proof, so all good --- can you elaborate where/why you are struggling?

If you are categorically minded, you can also pursue a less 'computational' proof by observing that both sides of the purported isomorphism are left adjoint to the constant simplicial set functor.

Answer 2

One approach is to think about adjoint functors:

You want to show that $\pi_0\circ|-|\cong\text{colim}$ as functors from simplicial sets to sets. You may know that the realiziation functor $|-|$ has a right adjoint $\text{Sing}X$ given by the singular simplicial set mapping $[n]$ to $\hom(\Delta^n,X)$. Now you might also observe that $\pi_0$ as a functor from spaces to sets has a right adjoint which is given by mapping a set $S$ to the discrete topological space on $S$. You may now see that the composition of the two right adjoints, i.e. the functor mapping a set $S$ to the singular simplicial set on its discrete topological space is actually just the simplicial set $[n]\mapsto S$, i.e. the constant functor. Now since adjoints compose we know that the composition of the left adjoint $\pi_0\circ|-|$ is left adjoint to the composition of the right adjoints, which is just the constant diagram functor. Now in general, taking colimits (if they all exist) is a left adjoint (here as a functor $\text{colim}\colon\text{Fun}(\Delta^{\text{op}},\text{Set})\to\text{Set}$) to the constant diagram functor. It then follows from the uniqueness (up to isomorphism) of adjoint functors,that $\pi_0\circ|-|\cong\text{colim}$.

Apart from avoiding explicit constructions on a set level, an advantage of this approach is that - assuming you know/prove all involved adjunctions - you get an isomorphism of functors. This means that the isomorphism in the question is natural in $X$.