Proposition 4.8.5.20 (05AF)

Proposition 4.8.5.20. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then:

$(a)$: The morphism $f$ is $0$-full (in the sense of Definition 4.8.5.10) if and only if the induced map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.
$(b)$: For $n \geq 1$, the morphism $f$ is $n$-full if and only if, for every vertex $x \in X$ having image $y = f(x)$, the induced map $\pi _{m}(f): \pi _{m}(X,x) \rightarrow \pi _{m}(Y,y)$ is injective for $m = n-1$ and surjective for $m = n$.

Proof. Assertion $(a)$ is immediate from the definitions. We will prove $(b)$. The case $n=1$ follows from Exercise 4.8.5.7. Let us therefore assume that $n \geq 2$. By definition, $f$ is $n$-full if and only if, for every edge $u: x \rightarrow x'$ of $X$ having image $v: y \rightarrow y'$ in $Y$, the induced map

\[ \pi _{m-1}( \operatorname{Hom}_{X}(x,x'), u) \rightarrow \pi _{m-1}( \operatorname{Hom}_{Y}( y, y'), v ) \]

is injective for $m = n-1$ and surjective for $m = n$. By virtue of Remark 4.8.5.19, it suffices to check this in the special case where $u = \operatorname{id}_{x}$ is a degenerate edge of $X$. Assertion $(b)$ now follows from isomorphisms

\[ \pi _{m-1}( \operatorname{Hom}_{X}(x,x), \operatorname{id}_{x} ) \simeq \pi _{m}(X,x) \quad \quad \pi _{m-1}( \operatorname{Hom}_{Y}(y,y), \operatorname{id}_{ y } ) \simeq \pi _{m}(Y, y) \]

of Example 4.6.1.13. $\square$