I understand that my question is probably elementary to someone well-versed in model categories, but the subject is very deep and I wonder whether there is a much simpler answer. 


If you localize a ring, some elements get identified, so I assume the same will happen when we localize a category.


Suppose we start with the category of pointed topological spaces and localize it at homotopy equivalences. I assume such thing exists (defined by a universal property) and I will treat it formally. 


I came up with the following argument that if $X$ is contractible, $End(X)$ is just the identity - if we compose two maps $f, g$ them with zero, they are the same, and since zero is invertible in $End(X)$, $f=g$. 


However, I'm having problems showing that, in general, if $f, g: X \rightarrow Y$ are homotopic, they are equal in the localization. Is it even true, ie. do we recover the homotopy category of topological spaces in the localization?


If not, then how much can we prove? What changes if I localize at weak equivalences? What happens if we limit ourselves to $CW-complexes$?



Localizing at homotopy equivalences
If we localize at all weak equivalences, the localization $L^W(Top)$ will be equivalent to the homotopy category of CW-complexes $Ho(CW)$. This can be seen using the fact that for each topological space there is a weakly equivalent CW-complex (the cellular approximation) and the fact that on CW-complexes the weak equivalences are exactly the homotopy equivalences (Whitehead theorem).


However, we can also localize the category of topological space at the homotopy equivalences. Then we will get the Homotopy category of all topological spaces $Ho(Top)$.


The fact, that if $f, g: X \to Y$ are homotopic, that they are equal in the localization, can be seen using the fact that the cylinder $X \times I$ used to define homotopy is isomorphic to $X$ in the homotopy category. This then shows that there is a functor


$F: Ho(Top) \longrightarrow L^W(Top) \cong Ho(CW)$


But there is of course also the inclusion $Ho(CW) \to Ho(Top)$ which is right inverse to $F$.


