For simplicial sets $X$ and $Y$, let us denote by $\underline{\mathrm{Hom}}(X, Y)$ the simplicial set of morphisms $X \to Y$. If $p: X \to Z$ and $q: Y \to Z$ are morphisms of simplicial sets, let us define the simplicial set of morphisms $X \to Y$ over $Z$ (or $p \to q$ in the slice, if you prefer) as the simplicial set $\underline{\mathrm{Hom}}(p, q)$ given by the pullback of the following corner:
$$\require{AMScd}
\begin{CD}
& &\underline{\mathrm{Hom}}(X, Y)\\
& @V{q_\ast}VV\\
\ast @>{p}>> \underline{\mathrm{Hom}}(X, Z) ;
\end{CD}$$
It is almost obvious that if $q$ is a Kan fibration, then $\underline{\mathrm{Hom}}(p, q)$ is a Kan complex, as the exponential of a Kan fibration is a Kan fibration, too. What I cannot prove is the following: if $i_A: A\hookrightarrow X$ is the inclusion of a simplicial subset and $q$ is a Kan fibration, then the restriction map $$ i_A^\ast : \underline{\mathrm{Hom}}(p, q) \to \underline{\mathrm{Hom}}(p|_A, q)$$ is a Kan fibration. Has anyone got any idea?
Best Answer
The Kan–Quillen model structure on simplicial sets satisifies axiom SM7, which means the canonical morphism $$[X, Y] \to [A, Y] \times_{[A, Z]} [X, Z]$$ induced by a monomorphism $A \to X$ and a Kan fibration $Y \to Z$ is a Kan fibration. Consider the following commutative diagram: $$\require{AMScd} \begin{CD} [p, q] @>>> [p \circ i, q] @>>> \Delta^0 \\ @VVV @VVV @VVV \\ [X, Y] @>>> [A, Y] \times_{[A, Z]} [X, Z] @>>> [X, Z] \\ & @VVV @VVV \\ & &[A, Y] @>>> [A, Z] \end{CD}$$ The wide rectangle and the tall rectangle are pullbacks by definition. The bottom right square is a pullback, so the top right square is a pullback, and it follows that the top left square is also a pullback. Hence, $[p, q] \to [p \circ i, q]$ is a pullback of $[X, Y] \to [A, Y] \times_{[A, Z]} [X, Z]$, so it is indeed a Kan fibration.