Proof:

The proof writes itself out. Suppose that $G:X\times [0,1]\longrightarrow Y$ is a homotopy between $f$ and the constant map taking the value $y_0$ say,

$$G(x, 1) = f(x),\quad G(x, 0) = f(y_0),$$    for all $$\;x\in X.$$

The second equation in (25.12) says that $G$ respects the identification made on $X\times [0,1]$ to yield $(X\times [0,1])/(X\times\{0\})$ whereby we conclude the existence of a map $F:C(X)\longrightarrow Y$ satisfying $F\circ \eta = G$. This map $F$ is continuous by the universal property and the first equation in (25.12) gives $F[x, 1] = G(x, 1) = f(x)$. The proof of necessity is complete.

Conversely suppose given a continuous map $f : X \longrightarrow Y$ such that there is a $G:C(X)\longrightarrow Y$ with $F\circ i = G$. Denoting by $\eta$ the quotient map $X\times [0,1]\longrightarrow C(X)$, the map $G\circ \eta$ provides a homotopy between $f$ the constant map. $\square$