Let $\mathbf{TOP}$ be the category of topological spaces and $\mathbf{HO(TOP)}$ be the category whose objects are topological spaces and morphisms are equivalence classes of continuous maps. We have a canonical functor $q: \mathbf{TOP}\rightarrow \mathbf{HO(TOP)}$ that sends an object to itself and a map to it's equivalence class.
Now I am trying to see that this map has the universal property that if $F:\mathbf{TOP}\rightarrow C$ is a functor that sends homotopy equivalences to isomorphisms, then it extends uniquely to a functor $F':\mathbf{HO(TOP)}\rightarrow C$ such that $F=F'\circ q$. Now to do this I need to show that if $f\cong f'$ then $F(f)=F(f')$.
Now I am bit lost on how to achieve this even for a simple case where I assume that $f\cong id$ then I would like to see tht $F(f)=id$. I know that since $f\cong id$ then $f$ is an homotopy equivalence and so $F(f)$ is an isomorphism and after playing around with it and $F(f^k)$ I was not able to see why I would have that $F(f)=id$.
Now I am not sure if I am confusing something or just forgetting to use some propery but any hint or help is appreciated. Thanks in advance.
Best Answer
Hint : consider the projection map $X\times [0,1]\to X$