Proof criticize: $Sing(X)$ is fibrant

Question

I am trying to prove the question in the title where definitions are given below.
Definition (fibration). A map $f:X\to Y$ in the category of simplicial sets is a fibration if it satisfies right lifting property with resepct to all inclusions $\Lambda^n_k\subseteq \Delta^n$ for $0\leq k\leq n >0$. And a simplicial set $X$ is fibrant if $X\to\{*\}$ is a fibration.
Definition (Sing(X)). Given a topological space $X$, we define the simplicial set $Sing(X)$ by $$Sing(X)([n])=Hom_{Top}(|\Delta^n|,X)$$ Here is my approach: given a commutative diagram in $sSet$ $$\require{AMScd} \begin{CD} \Lambda^n_k @>{\sigma}>> Sing(X);\\ @VV{i}V @VVV \\ \Delta^n @>{}>> \{*\}; \end{CD}$$ I apply the realization functor $|\cdot|:sSet\to Top$ to get a commutative diagram $$\require{AMScd} \begin{CD} |\Lambda^n_k| @>{|\sigma|}>> |Sing(X)|;\\ @VV{|i|}V @VVV \\ |\Delta^n| @>{}>> |\{*\}|; \end{CD}$$ Now since there is a retraction $r:|\Delta^n|\to |\Lambda^n_k|$ such that $r|i|=id_{\Lambda^n_k}$, I define the lifting in the new commutative diagram by $|\sigma|r$. Now I wish to go back to $sSet$ but I am not sure if I am on the right track.
A naural idea of mine is to apply $Sing$ again, then extend the diagram on the left hand side by the unit map, and extend on the right hand side by the counit map. This seems to work. $$\begin{array}{} \Lambda^n_k &\rightarrow & Sing|\Lambda^n_k|& \rightarrow & Sing|Sing(X)| &\rightarrow &Sing(X)\\ \downarrow&&\downarrow&\nearrow&\downarrow&&\downarrow&\\ \Delta^n &\rightarrow & Sing|\Delta^n|& \rightarrow & Sing|*| &\rightarrow &* \end{array}$$ But it feels weird. Is it correct? Or is there a simplier way of seeing this? Thanks in advance.

Answer 1

This is the correct approach, but it may be clarifying to see it happen in its proper generality. A (somewhat careful, if needed) general categorical argument shows the following: given an adjoint pair $F\colon\mathcal{C}\rightleftarrows\mathcal{D}\colon G$ of functors, a map $i\colon A\to B$ in $\mathcal{C}$ and a map $p\colon X\to Y$ in $\mathcal{D}$, a square $$ \require{AMScd} \begin{CD} A @>>> GX\\ @V{i}VV @VV{Gp}V\\ B @>>> GY \end{CD} $$ admits a lift $B\to GX$ in $\mathcal{C}$ iff the corresponding ''adjoint'' square $$ \require{AMScd} \begin{CD} FA @>>> X\\ @V{Fi}VV @VV{p}V\\ FB @>>> Y \end{CD} $$ admits a lift $FB\to X$ in $\mathcal{D}$, and when they exist the lifts $B\to GX$ and $FB\to X$ correspond to one another under the adjunction $F\dashv G$. This may be fun to show for yourself, but if you want to see a proof, let me know.

Using that $\mathrm{Sing}(*)\cong\Delta^0$, you can use your ''topological'' lifts to obtain ''simplicial'' lifts, and recognize your approach as a special case of the categorical statement above (which is worth remembering).