I am stuck with proving the following:

For each topological space $X$, the singular simplicial set $\mathrm{Sing}_\bullet(X)$ is a Kan complex.

By the definition of Kan complex, we need to do the following: given any $n>0$, $0\leq i \leq n$, and $\sigma_0\colon \Lambda^n_i\to\mathrm{Sing}_\bullet(X)$, find a $\sigma\colon \Delta ^n \to\mathrm{Sing}_\bullet(X)$ such that $\sigma_0$ factors as $\Lambda^n_i\hookrightarrow\Delta^n\to \mathrm{Sing}_\bullet(X)$, where the last arrow is $\sigma$.

An obvious approach seems to use the adjunction $|-|\dashv \mathrm{Sing}_\bullet$ between the geometric realization functor and the singular simplicial set functor to translate $\sigma_0\colon \Lambda^n_i\to\mathrm{Sing}_\bullet(X)$ into a continuous map $f_0\colon |\Lambda^n_i|\to X$. Then we can find an $f\colon |\Delta^n|\to X$ such that $f_0$ factors as $|\Lambda^n_i|\hookrightarrow |\Delta^n|\to X$, where the last arrow is $f$. For instance, let $\ell$ be a continuous left inverse of the inclusion $|\Lambda^n_i|\hookrightarrow|\Delta^n|$. Then set $f:=f_0\circ \ell$.

Now, using the adjunction $|-|\dashv \mathrm{Sing}_\bullet$, we can translate the continuous map $f\colon |\Delta^n|\to X$ into a simplicial map $\sigma\colon\Delta^n\to\mathrm{Sing}_\bullet(X)$.

Question: Why does $\sigma\colon\Delta^n\to\mathrm{Sing}_\bullet(X)$ have the property that it factors as $\Lambda^n_i\hookrightarrow\Delta^n\to \mathrm{Sing}_\bullet(X)$? I guess we can't just use the fact that there merely exists a natural bijection $\hom(|S_\bullet|, X)\cong\hom(S_\bullet, \mathrm{Sing}_\bullet)$, but we have to use concrete properties of a specific one, right? Otherwise I don't see how one could conclude that property, if $\sigma$ is just the image of some such bijection.


1 Answer 1


This follows immediately from naturality of the bijection $\alpha_{S_\bullet}:\hom(|S_\bullet|, X)\cong\hom(S_\bullet, \mathrm{Sing}_\bullet(X))$. Specifically, naturality of that bijection with respect to the inclusion $j:\Lambda^n_i\to\Delta^n$ says the following diagram commutes:

$$\require{AMScd} \begin{CD} \hom(|\Delta^n|, X) @>{\alpha_{\Delta^n}}>> \hom(\Delta^n, \mathrm{Sing}_\bullet(X)) \\ @V{|j|^*}VV @V{j^*}VV \\ \hom(|\Lambda^n_i|, X) @>{\alpha_{\Lambda^n_i}}>> \hom(\Lambda^n_i, \mathrm{Sing}_\bullet(X)) \end{CD}$$ You have found $f\in \hom(|\Delta^n|, X)$ such that $|j|^*(f)=f_0=\alpha_{\Lambda^n_i}^{-1}(\sigma_0)$. Commutativity of the digram then says that $j^*(\alpha_{\Delta^n}(f))=\sigma_0$, i.e. that your $\sigma$ satisfies $\sigma\circ j=\sigma_0$, as desired.

  • $\begingroup$ That's cool, thanks! So we can solve the extension property in the category of topological spaces and then translate the solution back to the category of Kan complexes. Can this trick of translating properties along an adjunction be generalized? Does it work for problems other than extension problems too? I know that functors always preserve (but not reflect) commutative diagrams. Is there a general statement about which types of statements are preserved by adjunctions? $\endgroup$
    – user984603
    Commented Oct 28, 2021 at 18:20
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    $\begingroup$ I wouldn't worry about any sort of single grand generalization--this is something you'll get a better intuition for just from experience with more examples. The one thing I would recommend to keep in mind is that if some equality of morphisms built from universal properties looks like it could be true (like $\sigma\circ j=\sigma_0$ in your question), it probably is true, and it's just a matter of unraveling all the functoriality and naturality statements to formally prove it. $\endgroup$ Commented Oct 28, 2021 at 18:24
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    $\begingroup$ Basically, if you have a statement that looks like it ought to be true by just interpreting the meanings of things like adjunctions "naively" without regard for the precise statement in terms of natural isomorphisms (e.g., saying that morphisms $|\Delta^n|\to X$ are "the same" as morphisms $\Delta^n\to Sing(X)$ and similarly for $\Lambda^n_i$, so they ought to factor through on one side when they do on the other side), it usually actually is true. $\endgroup$ Commented Oct 28, 2021 at 18:27
  • $\begingroup$ These comments are very helpful. I couldn't stop worrying about whether there is such a generalization, though. Are you aware of any? I asked a follow-up question here: math.stackexchange.com/questions/4300519/… $\endgroup$
    – user984603
    Commented Nov 8, 2021 at 18:01

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