Warning 3.1.7.3. In the situation of Corollary 3.1.7.2, the Kan complex $Q$ is not uniquely determined. However, the homotopy type of $Q$ depends only on $X$. If $Q'$ is another Kan complex equipped with a map $f': X \rightarrow Q'$, then we can write $f' = g \circ f$ for some map of Kan complexes $g: Q \rightarrow Q'$ (Remark 3.1.2.7). If $f'$ is a weak homotopy equivalence, then $g$ is also a weak homotopy equivalence (Remark 3.1.6.16) and therefore a homotopy equivalence (Proposition 3.1.6.13).
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$