I know that the loop space of given pointed topological space $(X,\ast)$ is the set of pointed maps $\mathrm{Map}_\ast(S^1,X)$. I would denote it by $\Omega X$.

On the other hand, I saw an article in $n$Lab which contains a categorical definition of loop space: In a $(\infty,1)$-category with a zero-object $0$, the loop space of object $X$ is the pullback of the diagram $0\rightarrow X\leftarrow 0$ (,as far as I have understood).

I have tried to apply this definition to the category $\mathrm{Top}_\ast$, which is a $(\infty,1)$-category (as $n$-morphisms are $n$-homotopies) with the zero-object $\{\ast\}$. However, the pullback of $\{\ast\}\rightarrow X\leftarrow\{\ast\}$ seems to be $\{\ast\}$, not $\Omega X$. And, even if we accept that $\Omega X$ is the pullback, then there have to be an up-to-homotopy unique pointed map $Y\rightarrow \Omega X$ for any pointed space $Y$, because two zero-maps $Y\rightrightarrows X$ are homotopic by definition. I cannot imagine such a map $Y\rightarrow \Omega X$ except the trivial zero-map. This is not desirable for me because originally I am interested in $\Omega$-spectrum and thus the map $Y\rightarrow \Omega X$ should be a homotopy equivalence. So I got confused completely.

Is not the categorical object defined as above the categorical counterpart of $\Omega X$? Is the explanation of the article a bit incorrect? Or have I misunderstood something?

(I glanced at a referenced paper by Lurie, but I couldn't read the construction because of my poor knowledge about category theory...)

  • 2
    $\begingroup$ $(\infty,1)$-categories have $(\infty,1)$-pullbacks, in general you should not expect them to have "ordinary" pullbacks $\endgroup$ – მამუკა ჯიბლაძე Nov 8 '17 at 4:53
  • $\begingroup$ That's the misunderstanding I made! $\endgroup$ – Y. S Nov 8 '17 at 5:09
  • 1
    $\begingroup$ Not sure if you have already understood this, but if you want to compute the homotopy pullback correctly, you can replace one of the $\{\ast\}$ with the path space $PX$ (paths starting at $\ast$). The path space is contractible, and $PX\to X$ (which sends a path to its endpoint) is a fibration. Thus the homotopy pullback can be computed as the naive fiber product of $PX \times_X \{\ast\}$ which is exactly $\Omega X$. The homotopy pullback models the pullback in the sense of $(\infty,1)$-categories. $\endgroup$ – Sam Gunningham Nov 8 '17 at 11:17
  • 1
    $\begingroup$ Another way of seeing it. is that a map into the homotopy pullback of $A\leftarrow B\rightarrow C$ is the same thing as a map to $A$, a map to $C$ and a homotopy of their postcompositions to $B$. Plugging in the diagram $*\to X\leftarrow *$ you see that a map to the homotopy pullback is a homotopy of the constant map to itself, that is a map to $\Omega X$. $\endgroup$ – Denis Nardin Nov 8 '17 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.