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.
Best Answer
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.