Let $f\colon X\to Y$ be a map of sets.
Proof of Proposition 2.5.4.1.3.
Item 1: Interaction With Functions
Indeed, we have
\begin{align*} \webleft [f_{*}\circ \chi _{X}\webright ]\webleft (x\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{*}\webleft (\chi _{X}\webleft (x\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{*}\webleft (\webleft\{ x\webright\} \webright )\\ & = \webleft\{ f\webleft (x\webright )\webright\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\chi _{X'}\webleft (f\webleft (x\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\chi _{X'}\circ f\webright ]\webleft (x\webright ),\end{align*}
for each $x\in X$, showing the desired equality.
