$\DeclareMathOperator\Ho{Ho}\DeclareMathOperator\Hom{Hom}$There are (at least) seven kinds of morphism spaces for a simplicial set $X$:

  1. The left-pinched morphism space $\Hom^L_X(x,y)$,
  2. The right-pinched morphism space $\Hom^R_X(x,y)$,
  3. The (non-pinched) morphism space $\Hom_X(x,y)$,
  4. The simplicial set $\Hom_{\mathfrak{C}[X]}(x,y)$ where $\mathfrak{C}[X]$ is the rigidification of $X$,
  5. The simplicial set $\Hom_{\mathfrak{C}^{nec}[X]}(x,y)$ where $\mathfrak{C}^{nec}[X]$ is the Dugger-Spivak rigidification of $X$,
  6. The simplicial set $\Hom_{\mathfrak{C}^{hoc}[X]}(x,y)$ where $\mathfrak{C}^{hoc}[X]$ is the simplicial set defined in page 17 of Dugger-Spivak's Rigidification of quasi-categories,
  7. The simplicial set $\Hom^E_X(x,y)$ defined in page 15 of Dugger-Spivak's Mapping spaces in Quasi-categories.

Do these mapping spaces all agree on the "underived level"? I.e. do we have $$ \pi_0\Hom^L_X(x,y)\cong\pi_0\Hom^R_X(x,y)\cong\pi_0\Hom_X(x,y) $$ $$\cong\pi_0\Hom_{\mathfrak{C}[X]}(x,y)\cong\pi_0\Hom_{\mathfrak{C}^{nec}[X]}(x,y)\cong\pi_0\Hom_{\mathfrak{C}^{hoc}[X]}(x,y) $$ $$ \cong\pi_0\Hom^E_X(x,y) $$ for an arbitrary simplicial set $X$?

In particular, is this true in the special case where $x=y$? (Such a restriction helps already for the left/right pinched morphism spaces: by Tag 01KZ, we have $\Hom^L_X(x,x)\cong\Hom^R_X(x,x)^\mathrm{op}$, and thus $\pi_0\Hom^L_X(x,x)\cong\pi_0\Hom^R_X(x,x)$ since $\pi_0(X^\mathrm{op})\cong\pi_0(X)$. Edit: this is wrong, see R. van Dobben de Bruyn's answer below)

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    $\begingroup$ I thought the first issue is that $\pi_n$ doesn't really make sense if $X$ isn't a Kan complex. Am I missing something? $\endgroup$
    – cheyne
    Nov 4, 2022 at 18:12
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    $\begingroup$ @cheyne The definition of the $\pi_n$ of a Kan complex does not make sense for a general simplicial set since homotopy can fail to be an equivalence relation, but $\pi_0$ works okay, being given by $\mathrm{colim}(X)\cong\mathrm{Coeq}(X_0\rightrightarrows X_1)$. The motivation I had in mind for this question is that one could perhaps define higher homotopy groups* for general simplicial sets by letting $\Omega X=\mathrm{Hom}_X(x,x)$, and then considering $\pi_0(\Omega\cdots\Omega X)$. (*They will be just sets in general) $\endgroup$
    – Emily
    Nov 4, 2022 at 18:29
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    $\begingroup$ The one big problem for this definition however would be if $\pi_0\mathrm{Hom}_X(x,x)$ didn't agree with $\pi_0\mathrm{Hom}^L_X(x,x)$ or $\pi_0\mathrm{Hom}^R_X(x,x)$. $\endgroup$
    – Emily
    Nov 4, 2022 at 18:31
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    $\begingroup$ (It's also worth mentioning that $\pi_0$ of the endomorphism spaces can behave in unexpected ways for general simplicial sets because of their "directedness", e.g. $\pi_1(\partial\Delta^2)$ defined as above is a point, while $\pi_1$ of the geometric realization of $\partial\Delta^2$ (which is a circle) is $\mathbb{Z}$ (this makes sense though: the $\pi_0$ of the category "that looks like" $\partial\Delta^2$ is again just a point.) $\endgroup$
    – Emily
    Nov 4, 2022 at 18:31
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    $\begingroup$ N.B. #4 and #5 are weak homotopy equivalent for every simplicial set $X$. (This is one of the main theorems of the Dugger--Spivak paper.) $\endgroup$ Nov 5, 2022 at 2:16

1 Answer 1


There is no hope of comparisons between $\pi_0\operatorname{Hom}^L_X(x,y)$ and $\pi_0\operatorname{Hom}^R_X(x,y)$ in general, even when $x=y$. (Note that the statement you cited says $\operatorname{Hom}^L_{X^{\text{op}}}(x,x) \cong \operatorname{Hom}_X^R(x,x)^{\text{op}}$, so it computes Homs in a different simplicial set.)

Example. Let $X$ be the coequaliser of $f,g \colon \Delta^0 \amalg \Delta^1 \rightrightarrows \Delta^2$, where $f$ maps $\Delta_0$ to $0$ and $\Delta^1$ to $\operatorname{id}_0$, and $g$ maps $\Delta_0$ to $2$ and $\Delta^1$ to the arrow $0 \to 1$. The nondegenerate simplices in $X$ are:

  • A single vertex $x$ (the image of $0$, $1$, and $2$ in $\Delta^2$).
  • Two arrows $\alpha$ and $\beta$ (the images of $0 \to 2$ and $1 \to 2$; the third arrow $0 \to 1$ has become identified with a degenerate $1$-simplex).
  • One $2$-simplex $h$ (the image of $0 \to 1 \to 2$), by definition a right homotopy $\alpha \stackrel\sim\to \beta$.

Thus we see that $\alpha$ and $\beta$ are right homotopic (hence homotopic in $\operatorname{Hom}_X(x,x)$ as well), but not left homotopic as there are no other nondegenerate $2$-simplices. $\square$

In particular, any of the definitions that are self-dual (e.g. $\operatorname{Hom}_{X^{\text{op}}}(x,y) = \operatorname{Hom}_X(y,x)^{\text{op}}$) cannot agree with $\pi_0\operatorname{Hom}^L_X(x,y)$ or $\pi_0\operatorname{Hom}^R_X(x,y)$ in general. For instance, if $\pi_0\operatorname{Hom}^L_X(x,y) = \pi_0\operatorname{Hom}_X(x,y)$ for all $X$, then \begin{align*} \pi_0\operatorname{Hom}_X^R(x,y) &= \pi_0\operatorname{Hom}_{X^{\text{op}}}^L(y,x)^{\text{op}} = \pi_0 \operatorname{Hom}_{X^{\text{op}}}(y,x)^{\text{op}} \\ &= \pi_0 \operatorname{Hom}_X(x,y) = \pi_0 \operatorname{Hom}_X^L(x,y), \end{align*} which we saw is false (again feel free to assume $x=y$).

I believe all the others are self-dual (but I'm not familiar with all of them), so at least they don't agree with $\pi_0\operatorname{Hom}_X^L(x,y)$ or its dual. I'm not sure if any subset of the others agree in general, but you can probably make counterexamples like the one above. (The convenient coincidence is that finite simplicial sets are far removed from the Kan or inner horn conditions: from some dimension on, you can only produce degenerate simplices, which do a bad job filling horns.)

  • $\begingroup$ Thank you so much, this is precisely what I was looking for! $\endgroup$
    – Emily
    Nov 5, 2022 at 21:32

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