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Given a topological space $X$, we may wish to consider it as an $(\infty,0)$-category, where the objects are the points of the space, the 1-morphisms are continuous paths between points, the 2-morphisms are homotopies of paths, etc. etc. This is just looking at the fundamental $\infty$-groupoid of the space $X$ (I think). Now, when we wish to work with such a category, is it typical to work up to ismorphism? That is, do we only think about the objects of our space in terms of their connected component? Is it "evil" to think otherwise? My questions can be clearly stated as follows:

1) Given a space $X$ regarded as an $(\infty,0)$-category, in practice, do we only work with connected components as our "objects" of the category?

2) In the case of the affirmative, I assume that all paths $p:x_1\to x_0$ such that $x_1\neq x_0$ are simply absorbed into one "class" of 1-morphisms represented by the identity morphism on the object corresponding to that component. However, do we have extra 1-morphisms which correspond to paths $p':x_0\to x_0$ which are not contractible?

3) In general, do we have one n-morphism corresponding to each "n-hole" of a space when we look at it as an $(\infty,0)$-category?

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I thought the $\infty$-point of view was to regard spaces as $(\infty, 0)$-categories, i.e. $\infty$-groupoids. Also, you seem to have confused isomorphisms and equivalences of categories... – Zhen Lin Jan 12 '12 at 23:07
How so? I am referring to isomorphisms within the category. – Jon Beardsley Jan 12 '12 at 23:14
And regarding your comment about the number I'm using, as far as I can tell is just notation. I mean that all $n$-morphisms for $n\geq 1$ are isomorphisms. I may be using the wrong notation. – Jon Beardsley Jan 12 '12 at 23:14
Upon reading a bit of this question may need to be reformulated or edited somewhat, i.e. after noting that homotopies must fix end points and so on and so forth. More will be revealed... – Jon Beardsley Jan 12 '12 at 23:26
The standard notation, as far as I know, is that a $(n, r)$-category has, for $k > r$ (strict inequality!), all $k$-morphisms isomorphisms. In particular, an ordinary category is a $(1, 1)$-category, and a preorder is a $(0, 1)$-category. – Zhen Lin Jan 12 '12 at 23:58
up vote 7 down vote accepted

I think Neil Strickland's suggestion to post the question on MO is reasonable; there are a lot more experienced homotopy theorists there. Nonetheless, I'll give it a shot.

Lurie's definition of the construction of an $(\infty, 1)$-category from a space is simply to take the singular simplicial set associated to it. This is a Kan complex, hence a quasicategory. (Conversely, any quasicategory where all edges are invertible is a Kan complex and can be thought of as representing a homotopy type.) The objects (or vertices of this simplicial set) are points of the space; the morphisms (or edges of this simplicial set) are maps from the unit interval into the space, etc. I don't think it makes sense to identify isomorphic vertices (i.e. vertices in the same connected component). However, one may ask that given a connected Kan complex, to produce a Kan complex with exactly one vertex which is homotopy equivalent to the first one.

In fact, there is a theory of minimal Kan complexes that answers this question (which could also be approached directly). Say that a Kan complex $X$ is minimal if whenever we have two $n$-simplices $x, y \in X_n$ which are homotopic relative to their boundaries (i.e. there exists a map $\Delta[n] \times \Delta[1] \to X$ restricting to $x,y$ and constant on $\partial \Delta[n] \times \Delta[1]$), we have $x = y$. It is a theorem then that any Kan complex contains a minimal Kan complex which is homotopy equivalent to the original one. So, we can use this to produce connected subKancomplexes of a connected Kan complex with precisely one vertex. (This is explained in ch. 1 of Goerss-Jardine, for instance.) One of the reasons this theory is useful is that minimal Kan complexes that are homotopic are actually isomorphic.

There is an analog of this theory for quasicategories (see HTT 2.3), which runs in a similar manner, except that when $n = 0$, the edge connecting the two should be an equivalence; then one gets the notion of a minimal quasicategory. It is similarly true that any quasicategory contains a minimal subquasicategory, to which it is categorically equivalent. I think the analogy in ordinary category theory of taking a minimal subquasicategory is the operation of taking a skeleton. Again, it turns out that categorically equivalent minimal quasicategories are isomorphic.

(A side comment: for reasons that I don't fully understand, the theory of minimal Kan fibrations (a slight generalization of the theory of minimal Kan complexes) plays an important role in the establishment of the usual model structure on simplicial sets. For instance, because a minimal Kan fibration is actually an honest (simplicial) fiber bundle, the geometric realization has to be a Serre fibration. Nonetheless, as far as I know, there is no such theory for quasi-categories. And the corresponding Joyal model structure seems to be much, much harder to establish -- in particular, no simple set of generating acyclic cofibrations are published in the literature, as far as I can tell! If someone knows how to do this, I would very much appreciate a pointer.)

Let me add that, as far as I understand, Lurie's theory of $(\infty, 1)$-categories (which is the only one I've looked at) makes very little reference to higher morphisms: there is a definite notion of an object (a vertex in the simplicial set), and a morphism (an edge). But higher morphisms are more or less treated as a black box, and as far as I can tell, the idea is not to separate them out but simply to work with mapping spaces between objects (which can be recovered from his theory). Instead, categorical notions are interpreted as statements of homotopy theory on the nerves, and those are then generalized to quasicategories.

Perhaps it's also worth pointing out that $(\infty, 1)$-category theory is supposed to generalize both classical homotopy theory and 1-category theory. That is, there is an imbedding $$\text{Kan complexes} \hookrightarrow (\infty, 1)-\text{categories}$$ whose image is supposed to be precisely the $(\infty, 0)$-categories, that is the $\infty$-groupoids. In Lurie's theory at least, this is true, and furthermore, a categorical equivalence between Kan complexes is precisely the same thing as a homotopy equivalence. So the $\infty$-category of a space holds precisely the information of its homotopy type (which is quite a bit stronger than holding its homotopy groups).

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