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Proposition Let $m$ and $n$ be integers and let $f: X \rightarrow Y$ be an $(m+n)$-connective morphism of Kan complexes. Let $B$ be a simplicial set of dimension $\leq m$, and let $A \subseteq B$ be a simplicial subset. Then the restriction map

\[ u: \operatorname{Fun}( B, X ) \rightarrow \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) \]

is $n$-connective.

Proof. Without loss of generality, we may assume that $m \geq 0$ (otherwise, our hypothesis guarantees that $A$ and $B$ are empty, so that $u$ is an isomorphism) and that $n \geq 0$ (otherwise, the conclusion that $u$ is $n$-connective is vacuous). Using Proposition, we can factor $f$ as a composition $X \xrightarrow {j} X' \xrightarrow {f'} Y$, where $j$ is anodyne and $f'$ is a Kan fibration. Since $Y$ is a Kan complex, the simplicial set $X'$ is a Kan complex (Remark, so $j$ is a homotopy equivalence of Kan complexes. We then have a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B,X) \ar [d]^{j \circ } \ar [r]^-{u} & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) \ar [d] \\ \operatorname{Fun}(B, X' ) \ar [r]^-{u'} & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) } \]

where the vertical maps are homotopy equivalences (see Proposition Consequently, to show that $u$ is $n$-connective, it will suffice to show that $u'$ is $n$-connective. We may therefore replace $f$ by $f'$, and thereby reduce to proving Proposition in the special case where $f$ is a Kan fibration. In this case, $u$ is also a Kan fibration (Theorem By virtue of Proposition, it will suffice to show that if $B'$ is a simplicial set of dimension $\leq n$ and $A' \subseteq B'$ is a simplicial subset, then every lifting problem

\begin{equation} \begin{gathered}\label{equation:exponentiation-for-connectivity} \xymatrix@R =50pt@C=50pt{ A' \ar [r] \ar [d] & \operatorname{Fun}( B, X ) \ar [d]^{u} \\ B' \ar [r] \ar@ {-->}[ur] & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) } \end{gathered} \end{equation}

admits a solution. Unwinding the definitions, we can rewrite (3.71) as a lifting problem

\[ \xymatrix@R =50pt@C=50pt{ (A \times B') \coprod _{ (A \times A' ) } (B \times A') \ar [r] \ar [d] & X \ar [d]^{f} \\ B \times B' \ar [r] \ar@ {-->}[ur] & Y. } \]

Since the simplicial set $B \times B'$ has dimension $\leq m+n$ (Proposition, the existence of a solution follows from our assumption that $f$ is $(m+n)$-connective (Proposition $\square$