Complete Segal spaces and quasi-categories are two common models for the theory of $(\infty,1)$-categories, and both are equipped with a natural notion of hom spaces. For complete Segal spaces, which in particular are simplicial spaces $X\colon \Delta^{\rm op} \to \mathcal{S}$, we think of $X_0$ as the space of object, $X_1$ as the space of morphism &c. Thus if $x,y\in X_{0,0}$, then the space of homs should be $x\times^h_{X_0} X_1 \times^h_{X_0} y$.

On the other hand, if $\mathcal{C}$ is a quasi-category, and $x,y\in \mathcal{C}_0$ are objects in $\mathcal{C}$, then the mapping space from $x$ to $y$ is defined (for example, there are other versions) by the rule ${\rm Map}_\mathcal{C}(x,y)_n=\left\{ \sigma\colon \Delta^n\times\Delta^1 \to \mathcal{C} \mid \sigma|_{\Delta^n\times\{0\}}\equiv x,\;\sigma|_{\Delta^n\times\{1\}}\equiv y \right\}$.

If $X$ is a complete Segal space, then we can associate to it a quasi-category $\mathcal{C}(X)$ via the rule $\mathcal{C}(X)_n:=X_{n,0}$, and this induces an equivalence between the $(\infty,1)$-category of complete Segal spaces and that of quasi-categories.

In particular, this functor should match up the hom spaces on both sides - we should have a homotopy equivalence ${\rm Map}_{\mathcal{C}(X)}(x,y)\simeq x\times^h_{X_0} X_1 \times^h_{X_0} y$ whenever $X$ is a complete Segal space and $x,y\in X_{0,0}$. Unfortunately, I'm a bit of a novice in higher category theory and don't know the literature well enough to know where to look for such a result.

So my question is the following: does anyone know where this equivalence of mapping spaces is explicitly proved (or possibly with ${\rm Map}$ replaced with any equivalent notion of mapping spaces for quasi-categories, such as ${\rm Map}^{\rm R}$ or ${\rm Map}^{\rm L}$)?


1 Answer 1


This might not be what you want, but you can go the other way around: to a quasicategory $C$ you can associate a Segal space via $NC: [n]\mapsto Fun(\Delta^n, C)^\simeq$, by which I mean the largest sub-Kan-complex of the simplicial hom from $\Delta^n$ to $C$.

Now this $N$ functor is an inverse for your $\mathcal C$ up to weak equivalence, and fo this one it is clear that $x\times_{NC_0} NC_1 \times_{NC_0} y$ is exactly $Map_C(x,y)$ (the maps $NC_1\to NC_0$ are fibrations, so these pullbacks are homotopy pullbacks). In fact, these are really isomorphic.

So if you know that mapping objects are insensitive to equivalences (namely, an equivalence of complete Segal spaces $f:X\to Y$ induces an equivalence $x\times^h_{X_0}X_1\times^h_{X_0} y\to f(x)\times^h_{Y_0}Y_1\times^h_{Y_0} f(y)$); and you know this claim that $N$ and $\mathcal C$ are weak inverses to one another, then for a given $X$ you can find equivalences $x\times^h_{X_0}X_1\times^h_{X_0} y \simeq x\times^h_{N\mathcal C X_0}N\mathcal C X_1\times^h_{N\mathcal C X_0} y\simeq Map_{\mathcal C(X)}(x,y)$

  • $\begingroup$ Ah brilliant, thanks! This does the trick perfectly. $\endgroup$
    – ChrisLazda
    Oct 4 at 16:42

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