Variant 3.5.9.19. Let $f: X \rightarrow Y$ be a morphism between Kan complexes, let $n$ be an integer, and let $k$ be a nonnegative integer. If $f$ is $n$-truncated, then the restriction map
is $(n-k)$-truncated. The converse holds if $k \geq n+2$.
Variant 3.5.9.19. Let $f: X \rightarrow Y$ be a morphism between Kan complexes, let $n$ be an integer, and let $k$ be a nonnegative integer. If $f$ is $n$-truncated, then the restriction map
is $(n-k)$-truncated. The converse holds if $k \geq n+2$.
Proof. Using Proposition 3.1.7.1, we can factor $f$ as a composition $X \xrightarrow {i} X' \xrightarrow {f'} Y$, where $i$ is anodyne and $f'$ is a Kan fibration. Then $X'$ is a Kan complex (Remark 3.1.1.11), so $i$ is a homotopy equivalence. We then have a commutative diagram
where the vertical maps are homotopy equivalences. It follows that $u$ is $(n-k)$-truncated if and only if $u'$ is $(n-k)$-truncated. We may therefore replace $f$ by $f'$ and thereby reduce to proving Variant 3.5.9.19 in the special case where $f$ is a Kan fibration. In this case, we have a commutative diagram
where the vertical maps are homotopy equivalences (by virtue of the contractibility of $\Delta ^{k}$). It follows that $u$ is $(n-k)$-truncated if and only if $\delta $ is $(n-k)$-truncated, so that Variant 3.5.9.19 is a reformulation of Corollary 3.5.9.16. $\square$