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Stream: theory: algebraic topology

Topic: trivial Kan fibrations

Leopold Schlicht (Dec 16 2021 at 17:36):

Look at this theorem. How can one deduce from that theorem that whenever $h$ and $h'$ are both composites of $f$ and $g$, then $h$ is homotopic to $h'$?

Recall that a simplicial set $X$ is called a contractible Kan complex if the unique map $X\to \Delta^0$ is a trivial Kan fibration. In the above theorem I interpret "contractible Kan complex" as something like "essentially a singleton", so that intuitively, the theorem says that the space of all composition witnesses $\sigma$ with $d_2\sigma = f$ and $d_0\sigma = g$ is essentially a singleton. (And intuitively, if such a $\sigma$ is unique, then in particular the composition $h=d_1\sigma$ of $f$ and $g$ is unique, which motivates the first question.)

Does that make sense: is a trivial Kan fibration something like a bijection, isomorphism, or equivalence between simplicial sets? If not, what is a trivial Kan fibration intuitively?

Let $G$ be a directed graph and $\mathcal C$ an $\infty$-category. This theorem says that there exists a trivial Kan fibration $\operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Path}[G] ), \mathcal{C}) \rightarrow \operatorname{Fun}( G_{\bullet }, \mathcal{C})$. ($\operatorname{Fun}$ denotes exponential objects, $\operatorname{Path}[G]$ is the path category of the directed graph $G$, and $G_\bullet$ is the $\leq 1$-dimensional simplicial set corresponding to $G$.) Does that imply that there is a bijection between the functors (simplicial maps) from $\operatorname{N}_{\bullet }( \operatorname{Path}[G])$ to $\mathcal C$ and the simplicial maps from $G_\bullet$ to $\mathcal C$? (A related fact I know is that an inner anodyne map is a bijection on vertices.)

Andrea Gentili (Dec 17 2021 at 17:58):

Leopold Schlicht said:

Look at this theorem. How can one deduce from that theorem that whenever $h$ and $h'$ are both composites of $f$ and $g$, then $h$ is homotopic to $h'$?

If you apply the definition of a trivial Kan fibration in the case $n=1$ to this situation, you obtain a map $\Delta^2\times\Delta^1\to\mathcal{C}$ constant on $\Lambda_2^1\times\Delta^1$ (that is, a homotopy between the compositions, seen as 2-simplices); in particuar, you obtain the homotopy between the composites you're looking for as the restriction along $d_1\times\mathrm{id}_{\Delta^1}\colon\Delta^1\times\Delta^1\hookrightarrow\Delta^2\times\Delta^1$.

Does that imply that there is a bijection between the functors (simplicial maps) from $\operatorname{N}_{\bullet }( \operatorname{Path}[G])$ to $\mathcal C$ and the simplicial maps from $G_\bullet$ to $\mathcal C$? (A related fact I know is that an inner anodyne map is a bijection on vertices.)

Not a bijection in general: consider the case $G_\bullet =\Lambda_2^1$, so that $\mathrm{Path}[G]=\Delta^2$ (as above, injectivity is up to homotopy).

Mike Shulman (Dec 17 2021 at 19:13):

A trivial Kan fibration is a special sort of equivalence.

Leopold Schlicht (Dec 27 2021 at 16:59):

@Andrea Gentili Thanks!

If you apply the definition of a trivial Kan fibration in the case $n=1$ to this situation, you obtain a map $\Delta^2\times\Delta^1\to\mathcal{C}$ constant on $\Lambda_2^1\times\Delta^1$ (that is, a homotopy between the compositions, seen as 2-simplices); in particuar, you obtain the homotopy between the composites you're looking for as the restriction along $d_1\times\mathrm{id}_{\Delta^1}\colon\Delta^1\times\Delta^1\hookrightarrow\Delta^2\times\Delta^1$.

What do you mean by "to this situation"? The only thing I can imagine is the following: let $\sigma$ be a witness of $h=g\circ f$ and $\sigma'$ be a witness of $h'=g\circ f$. Since $\operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}( \Lambda ^2_1, \operatorname{\mathcal{C}}) } \{ (g, \bullet , f) \}$ is a contractible Kan complex, every map $\partial \Delta^1\to \operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}( \Lambda ^2_1, \operatorname{\mathcal{C}}) } \{ (g, \bullet , f) \}$ can be extended to a map $\Delta^1\to \operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}( \Lambda ^2_1, \operatorname{\mathcal{C}}) } \{ (g, \bullet , f) \}$. Now we could plug in $(\sigma, \sigma')$ as the map $\partial \Delta^1\to \operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}( \Lambda ^2_1, \operatorname{\mathcal{C}}) } \{ (g, \bullet , f) \}$. But I don't see how a map of the type $\Delta^2\times\Delta^1\to\mathcal{C}$ appears.

@Mike Shulman Thanks. Can you make that more precise?

Mike Shulman (Dec 27 2021 at 17:07):

Any trivial Kan fibration is a simplicial homotopy equivalence. See for instance Corollary 1.4.5.5 here.

Leopold Schlicht (Dec 27 2021 at 17:10):

I didn't know that that's what is called a "simplicial homotopy equivalence". Thanks. I interpret this corollary as saying that a trivial Kan fibration is (surjective) and (injective up to homotopy) (as Andrea already said).

Andrea Gentili (Dec 27 2021 at 18:55):

You understood correctly. Once you have the map $\Delta^1\to\mathrm{Fun}(\Delta^2 ,\mathcal C)\times_{\mathrm{Fun}(\Lambda_2^1 ,\mathcal C)}\{ (g,\bullet ,f)\}$, you have to keep in mind that $\mathrm{Fun}(\Delta^2 ,\mathcal C)\times_{\mathrm{Fun}(\Lambda_2^1 ,\mathcal C)}\{ (g,\bullet ,f)\}$ can be seen as a subspace of $\mathrm{Fun}(\Delta^2 ,\mathcal C)$, so that you have a map $\Delta^1\to\mathrm{Fun}(\Delta^2 ,\mathcal C)$, that, by adjunction, gives you the map $\Delta^2\times\Delta^1\to\mathcal C$.

Leopold Schlicht (Dec 27 2021 at 19:34):

Thanks! I now see how to get the map $\Delta^1\times\Delta^1\hookrightarrow\Delta^2\times\Delta^1\to\mathcal C$. Are you then using Corollary 1.3.3.7 to obtain a homotopy $h\to h'$ from that map $\Delta^1\times\Delta^1\to\mathcal C$?

Leopold Schlicht (Dec 27 2021 at 19:45):

Mike Shulman said:

Any trivial Kan fibration is a simplicial homotopy equivalence. See for instance Corollary 1.4.5.5 here.

Ah, here is that statement. Also, if both the domain and the codomain of a trivial Kan fibration are $\infty$-categories, then the trivial Kan fibration is an equivalence of $\infty$-categories.

Andrea Gentili (Dec 27 2021 at 19:55):

Are you then using Corollary 1.3.3.7 to obtain a homotopy $h\to h'$ from that map $\Delta^1\times\Delta^1\to\mathcal C$?

Yes (the conditions of the corollary are satisfied since the original map factorizes through $\mathrm{Fun}(\Delta^2,\mathcal C)\times_{\mathrm{Fun}(\Lambda_2^1,\mathcal C)}\{ (g,\bullet ,f)\}$).

Leopold Schlicht (Dec 28 2021 at 10:39):

Thanks!