Prove that $X\mapsto CX$ defines a functor $Top\to Top$ (defining the behavior on morphisms). Hint use exercise 1.11.
Exercise 1.11. Let $X$ and $Y$ be spaces with equivalence relations $\sim$ and $\square$, respectively, and let $f: X \to Y$ be a continuous map preserving the relations (if $x\sim x'$, then $f(x) \square f(x')).$ Prove that the induced map $\overline f: X/\sim \to Y/\square$ is continuous; moreover, if $f$ is an identification, then so is $\overline f.$
In my research I found on Wikipedia that the map $X\to CX$ induces a functor $C:Top\to Top$.
If $f:X\to Y$ is a continuous map, then $Cf:CX\to CY$ is defined as $(Cf)([x,t])=[f(x),t]$
Is this the map that I should consider for the exercise? I would say no because the map in the exercise goes from $X$ only and goes to $CX$, whereas Wikipedia map goes from $CX$ to $CY.$
Thank you.
You may have some confusions here. A functor $C:\mathrm{Top} \to \mathrm{Top}$ is defined on objects but also on morphisms.
That is, you are given $C(X)=CX$, which tells you what to do on objects but you also need to know what it does on morphisms. In particular, given a continuous map $f:X \to Y$ (a morphism in $\mathrm{Top}$), we can define $C(f):C(X) \to C(Y)$, given by $C(f)([x,t])=[f(x),t]$.
Now, you have to show that this mapping satisfies the axioms of a functor. That is, given $f:X \to Y$, $id:X \to X$, and $g:Y \to Z$, then we need $C(id)=id:C(X) \to C(x)$ and also that $C(g \circ f)=C(g) \circ C(f)$.
Once all of this is done, you will have shown that $C$ is a functor.