I am currently reading Lurie's - Higher Topos Theory, but am having difficulties from the very beginning. I know that the topic is too broad and different mathematicians may define differently some of the notions that shall be mentioned in this post. Therefore I adapt the notation and terminology, Lurie adapts in his book. In view of that, he defines an $\infty-$category, to be a simplicial set say, $S \in Set_{\Delta}$, which satisfies the horn-condition, i.e., for any $n \geq 0$ and $0 < i < n$, any map $\Lambda^n_i \rightarrow S$, extends to a map $\Delta^n \rightarrow S$. However, some pages above, he explicitly says

"In this book, we consider only ($\infty$,1)-categories: that is, higher categories in which all k-morphisms are assumed to be invertible for $k > 1$."

While later on he writes out

Unless we specify otherwise, the generic term “$\infty$-category” will refer to an ($\infty$,1)-category.

Here is the point where I got stuck. In my understanding so far, these $\infty$-categories are literally functors, $\Delta^{op} \rightarrow Set$, with some additional "descent" condition to exploit the combinatorial nature of them. Moreover, for convenience we think of these $\infty$-categories as kind of generalisation of the usual categories, hence someone might think the $S_0$ as the objects, $S_1$ as the morphisms, $S_2$ as the triabgles and so on we can have in mind the usual nerve of a category as a compass to obtain some intuition. What I don't understand, is where the first quote comes into the stage. What does it mean to have $k$-morphisms in the $\infty$-category setup? And if so, what does it mean to have invertible morphisms in that sense?

Thanks in advance!


The intuitive idea of an $\infty$-category is a category-like structure where you have morphisms between morphisms between morphisms between morphisms and so on. That is, you have ordinary objects and morphisms, but you also have "2-morphisms" between parallel 1-morphisms (think natural transformations between functors, or homotopies between maps), and "3-morphisms" between parallel 2-morphisms, and so on. An $(\infty,1)$-category is then a structure of this sort where all the $k$-morphisms for $k>1$ are invertible. The basic motivating example is the $(\infty,1)$-category of spaces, where objects are (nice) topological spaces, morphisms are continuous maps, 2-morphisms are homotopies between maps, 3-morphisms are homotopies between homotopies, 4-morphisms are homotopies between homotopies between homotopies, and so on. Since every homotopy has an inverse (just reverse it), all the $k$-morphisms for $k>1$ are invertible.

So, Lurie's definition of "$\infty$-category" is actually just modeling this notion of $(\infty,1)$-category, not more general $\infty$-categories in which you can have non-invertible morphisms of all dimensions. You should think of an element of $S_2$ as not just a commuting triangle in the ordinary categorical sense, but a triangle which commutes up to a given homotopy. That is, you have objects $A$, $B$, and $C$, maps $A\to B$, $B\to C$, and $A\to C$, and a $2$-morphism ("homotopy") between the composition $A\to B\to C$ and the morphism $A\to C$. Similarly, higher-dimensional simplices are diagrams which commute up to given higher-dimensional morphisms.

In the case that you have an ordinary category, you consider it as an $(\infty,1)$-category by saying there are no $k$-morphisms for $k>1$ other than identity morphisms. So in this case all the higher morphisms in our simplices are identities, and the diagrams actually literally commute. So in that case, the simplicial set is just the usual nerve of the category.

  • $\begingroup$ Thanks for your response Eric. So if I understand correctly: in the Lurie's definition of an $\infty$-category, these $k$-morphisms are the diagrams of $S_k$, but interpreted them as a way to relate the $S_{k-1}$ diagrams. Those of the latter that can be related by an $S_k$ diagram are called homotopic, am I right? If so, how do we know that the related $(k-1)$-diagrams are invertible? In what sense? $\endgroup$ – user430191 Feb 22 '18 at 19:07
  • $\begingroup$ Well, the $k$-morphisms are related to the elements of $S_k$, but it's complicated, because the elements of $S_k$ are "simplex-shaped" and general $k$-morphisms have a different ("globular") shape. But it turns out that general $k$-morphisms can be "built out of" simplicial $k$-morphisms, and so Lurie's definition still gives a good notion of $(\infty,1)$-categories. $\endgroup$ – Eric Wofsey Feb 22 '18 at 19:18
  • $\begingroup$ As for why the $k$-morphisms are invertible for $k>1$, this is closely related to the horn condition. Note that if you require all horns to be filled, not just inner horns, then this gives you inverses for your morphisms (see Remark in Lurie's book). It turns out that being able to fill inner horns similarly gives you inverses for the $k$-morphisms for $k>1$. $\endgroup$ – Eric Wofsey Feb 22 '18 at 19:27
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    $\begingroup$ Here is one way to formulate it. Say $A$ and $B$ are two objects in an $\infty$-category $S$ (i.e., elements of $S_0$). You can defined the $\infty$-category $Hom_S(A,B)$ of morphisms between them as simplicial set of maps (using the usual internal hom of simplicial sets) $\Delta^1\to S$ which send the first vertex to $A$ and the second vertex to $B$. It turns out that the internal horn condition on $S$ implies that $Hom_S(A,B)$ fills all horns, not just internal ones (this is just some computation you can do using the combinatorics of simplicial sets). $\endgroup$ – Eric Wofsey Feb 22 '18 at 22:50
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    $\begingroup$ So that means that when you think of $Hom_S(A,B)$ as an $\infty$-category of its own, it is a groupoid (as in Remark the morphisms have inverses. But the morphisms in $Hom_S(A,B)$ are morphisms between morphisms from $A$ to $B$ in $S$: that is, they are 2-morphisms in $S$. In this way, the 2-morphisms in $S$ are all invertible. Similarly, the 3-morphisms in $S$ can be viewed as morphisms in an $\infty$-category of the form $Hom_{Hom_S(A,B)}(f,g)$, which are again invertible since this $\infty$-category will fill all horns. $\endgroup$ – Eric Wofsey Feb 22 '18 at 22:53

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