# Homotopy category of a simplicial category

In many places (for example here) I've seen the following definition:

For a simplicial category $\mathcal{C}$, it's homotopy category is defined to be the category $Ho(\mathcal{C})$ with the same objects as $\mathcal{C}$ and with morphisms $$Ho(\mathcal{C})(x,y)=\pi_0\mathcal{C}(x,y).$$ The problem I have with this definition is that homotopy groups of simplicial sets (in particular $\pi_0$) are well-defined only for Kan complexes.

So why does this definition work? Is $\mathcal{C}(x,y)$ necessarly a Kan complex for every $x,y\in \mathcal{C}$? Do we take fibrant replacement without loss of generallity?

• btw for "good" simplicial categories (fibrant ones wrt to the canonical model structure), the simplicial sets $C(x,y)$ will actually be Kan complexes so you can pick your favorite definition of $\pi_0$ – crystalline Feb 15 '16 at 17:33

## 1 Answer

$\pi_0$ is well-defined for any simplicial set $S$: it's the coequalizer of the two face maps $S_1 \to S_0$. This is the left adjoint to the inclusion of sets into simplicial sets.

• Oh ok. The only definition I've seen is that $\pi_0X$ is the set of homotopy class of maps $\Delta^0\to X$. How do these two defintion relate? Are they equivalent if $X$ is fibrant? – Nitrogen Feb 15 '16 at 4:16
• @Nitrogen: probably they are equivalent if $X$ is fibrant, but I wouldn't know. – Qiaochu Yuan Feb 15 '16 at 4:18
• It seems that they are indeed equivalent in the fibrant case. Thanks for your help! – Nitrogen Feb 15 '16 at 4:27
• I suppose it depends on what you mean "homotopy class" but under any reasonable interpretation it should be true. (For example, you might define it as a morphism in the homotopy category.) – Zhen Lin Feb 15 '16 at 13:57