MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $CW-complexes$?

share|cite|improve this question
So, what's the domain of a homotopy? // Incidentally, this particular localization is exactly what is usually meant by "the homotopy category". – some guy on the street Aug 16 '11 at 16:09
up vote 6 down vote accepted

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 $F$.

share|cite|improve this answer

I'm not sure if I'm reading your question correctly. Are you asking for a proof of the following statement?

If $f, g \colon X \to Y$ are homotopic map of topological spaces, then they are identified in the localization of the category of spaces with respect to homotopy equivalences.

The proof is completely elementary. Let $i_0, i_1 \colon X \to X \times I$ be the bottom and top inclusions of the cylinder on $X$. They admit a common homotopy inverse, namely the projection $X \times I \to X$ (it is in fact strong deformation retraction). Thus $i_0$ and $i_1$ are identified in the localization. If $H \colon X \times I \to Y$ is a homotopy from $f$ to $g$, then $f = H i_0$ and $g = H i_1$, so $f$ and $g$ are also identified.

In a general model category $\mathcal{C}$, if you assume that $X$ is cofibrant and $Y$ is fibrant, essentially the same argument tells you that $\mathrm{Ho}\mathcal{C}(X, Y)$ is $\mathcal{C}(X, Y)$ divided by the relation of left homotopy (or equivalently right homotopy). Of course you need to know that under (co)fibrancy assumptions every morphism in $\mathrm{Ho}\mathcal{C}(X, Y)$ is representable by a single arrow in $\mathcal{C}$ (as opposed to a longer zig-zag). In case of topological spaces and homotopy equivalences you don't need to do it, since every space is both cofibrant and fibrant in the Strøm model structure.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.