# Universal Property of the functor $q:\mathbf{TOP}\rightarrow \mathbf{HO(TOP)}$

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.

Hint : consider the projection map $$X\times [0,1]\to X$$
• I think I see how. We will have that $\pi$ is an homotopy equivalence and we can consider the inclusion $i_i: X\rightarrow X\times [0,1]$ such that $i_i(x)=(x,i)$ for $i=0,1$. Now we will have that if $f$ and $g$ are homotopic then $F(f)=F(H)\circ F(i_0)$ and $F(g)=F(H)\circ F(i_1)$ and using the uniquess of the inverse of $F(\pi)$ and the fact that $F(\pi \circ i_i)=F(id)=id$ we get that $F(i_0)=F(i_1)$ and so we get the desired result. @Maxime Ramzi. What do you think ? – Lost Mar 16 at 17:56