# Why are finite cell complexes also finite as infinity-categories?

A quasicategory ($$\infty$$-category) $$\mathcal{C}$$ is finite if there is a finite simplicial set $$K$$ and a categorical equivalence $$K\rightarrow\mathcal{C}$$.

On the other hand, a Kan complex (space) $$X$$ is finite if there is a finite simplicial set $$K$$ and a weak homotopy equivalence $$K\rightarrow X$$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes.

Question: Now suppose $$X$$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $$X$$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof?

To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $$K$$ and a weak homotopy equivalence $$f:K\rightarrow S^1$$. But $$f$$ is not a categorical equivalence. To build a finite model for the $$\infty$$-category $$S^1$$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other.

## 1 Answer

Start from a finite simplicial set $$K$$ which is homotopicaly equivalent to a Kan complex $$X$$.

Then by applying a finite number of pushout of outer horn inclusion to $$K$$, you can build homotopy equivalences $$K \hookrightarrow K' \rightarrow X$$ such that all the $$1$$-cells of $$K'$$ are "invertible" (in the sense that "for all $$1$$-cell $$f$$ there exists $$2$$-cells attesting homotopies $$g \circ f => 1$$ and $$f \circ h => 1$$ " see the "edit" below though ). $$K'$$ is still a finite simplicial set.

I claim that $$K' \rightarrow X$$ is now an equivalence in the Joyal model structure, which conclude the proof.

Indeed, as all the $$1$$-cells of $$K'$$ have homotopy inverses, the homotopy category of $$K'$$ (in the sense of the left adjoint to the nerve functor) is a groupoid.

So if I take $$K' \hookrightarrow Y \rightarrow X$$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $$Y$$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $$K'$$, hence is a groupoid, hence $$Y$$ is a Kan complex.

And $$Y \rightarrow X$$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $$K' \rightarrow X$$ is a Joyal equivalence.

Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:

For each $$1$$-cell of $$K$$ you use a pushout by a $$\Lambda^0 [2] \hookrightarrow \Delta[2]$$ and one by a $$\Lambda^2 [2] \hookrightarrow \Delta[2]$$ to add a cells $$g$$ and $$h$$ with $$2$$-cells $$f \circ g => 1$$ and $$h \circ f => 1$$.

And you stop there, we don't add any new cells (no right inverse for $$g$$, or left inverse for $$h$$)

This is enough to ensure that the homotopy category of $$K'$$ is a groupoids: the original cells of $$K$$ , like $$f$$, will be invertible because they have both a left inverse and a right inverse, and the new cells $$g$$ and $$h$$ are invertible because they are either right or left inverse to an invertible cell.

As every arrow the homotopy category of $$K'$$ is a composite of $$1$$-cell of $$K'$$ they will all be invertible.

• There is something I don't understand. Start with the simplicial set $\Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^\infty$, which is not finite. – John Berman Nov 6 '18 at 15:37
• @JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end. – Simon Henry Nov 6 '18 at 15:49