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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.

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Hint : consider the projection map $X\times [0,1]\to X$

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  • $\begingroup$ 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 ? $\endgroup$ – Lost Mar 16 at 17:56
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    $\begingroup$ That's exactly it, well done ! $\endgroup$ – Maxime Ramzi Mar 16 at 18:04

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