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Let $f:X \to Y$ be a homotopy equivalence of pointed topological spaces.

Then, is the induced map of pointed loop spaces $\Omega (f): \Omega X \to \Omega Y$ a homotopy equivalence?

Here, loop spaces are equipped with the compact-open topologies.

Is there any counterexample?

I do not know even whether the induced map $\Omega (f)$ is continuous or not in general.

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closed as off topic by Martin Brandenburg, Dan Petersen, Eric Wofsey, Oscar Randal-Williams, Fernando Muro Mar 10 '13 at 22:11

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The answer is yes always, provided $f\colon\thinspace (X,x_0)\to (Y,y_0)$ is a pointed homotopy equivalence of pointed spaces (meaning that the homotopies $g\circ f \simeq 1_X$ and $f\circ g\simeq 1_Y$ preserve the base points).

This follows from the fact that $\Omega$ is a homotopy functor from based spaces to based spaces, meaning in particular that

$$\Omega(g)\circ\Omega(f) = \Omega(g\circ f) \simeq \Omega(1_X) = 1_{\Omega X}$$


$$\Omega(f)\circ\Omega(g) = \Omega(f\circ g) \simeq \Omega(1_Y) = 1_{\Omega Y}.$$

To see that $\Omega$ is a homotopy functor, note that $\Omega(f)$ takes a loop $\gamma\colon\thinspace I\to X$ to the composition $f\circ \gamma\colon\thinspace I\to Y$. So if $F_t\colon X\to Y$ is a pointed homotopy from $f$ to $f'$, then $\Omega(F_t)\colon\thinspace \Omega(X)\to \Omega(Y)$ is a pointed homotopy from $\Omega(f)$ to $\Omega(f')$.

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Thanks for your comment. I wonder how one can prove that $\Omega (F_{t})$ is actually a homotopy. Is it always continuous without any assumption such as locally compactness? – Hiro Mar 10 '13 at 18:53
Actually, it is also straightforward to show that the map induced by any homotopy equivalence $f:X\to Y$ on based loop spaces is a homotopy equivalence. Simply observe that if $g:Y\to X$ is a homotopy inverse to $f$, then $(\Omega_{f(x)} g) \circ(\Omega_x f)$ is homotopic to "conjugation" by the path $H(x,-)$ for any homotopy $H:\mathrm{id}_X \simeq g\circ f$. Moreover, this "conjugation" map is a homotopy equivalence, since conjugation by the reverse path is a homotopy inverse. Similarly, $(\Omega_{g(f(x))} f) \circ(\Omega_{f(x)} g)$ is a homotopy equivalence. (to be continued) – Ricardo Andrade Mar 10 '13 at 23:23
(continuation) Since $(\Omega_{f(x)} g)\circ(\Omega_x f)$ is a homotopy equivalence, we conclude that $\Omega_{f(x)} g$ has a homotopy right inverse. Analogously, $\Omega_{f(x)} g$ has a homotopy left inverse because $(\Omega_{g(f(x))} f)\circ(\Omega_{f(x)} g)$ is a homotopy equivalence. Consequently, $\Omega_{f(x)}g$ is a homotopy equivalence. Finally, since $(\Omega_{f(x)} g)\circ(\Omega_x f)$ and $\Omega_{f(x)} g$ are both homotopy equivalences, so is $\Omega_x f$. This argument is a topological version of the usual proof that homotopy equivalences induce isomorphisms on fundamental groups. – Ricardo Andrade Mar 10 '13 at 23:33

Is the problem that you want to allow "pathological" topological spaces? I would have thought that for nice spaces all of this were sort of true by definition. (after all, $\Omega(X)=\bullet \times^h_X \bullet$ and homotopy limits are invariant under weak equivalences, no?)

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Thanks for your comment. I do not know anything about homotopy limits, but how about if the spaces are CW complexes? – Hiro Mar 10 '13 at 18:56
CW complexes are the opposite of "pathological". – Aaron Mazel-Gee Mar 10 '13 at 19:12
but definitely understanding why the homotopy fibre product $\bullet \times_X^h \bullet$ is the (based) loopspace is worth the time (~30 seconds). It's such a cool thing! – Jacob Bell Mar 10 '13 at 19:13
This is a fairly overpowered answer to a rather simple question... By the way, the homotopy invariance (with respect to homotopy equivalences) of homotopy pullbacks holds for all spaces (not just "nice" spaces), and is a fairly elementary exercise. Even more generally, homotopy limits of spaces preserve homotopy equivalences. This is a consequence of the homotopy invariance of homotopy limits in model categories applied to the Strøm model structure (see, together with the fact that all objects are fibrant in the Strøm model structure. – Ricardo Andrade Mar 11 '13 at 1:34
@ricardo: I agree, I put it out there as, to me, this perspective sweeps under the rug the actual maths and makes things look formal and easy. I did say in my second comment that I was cheating. :) – Jacob Bell Mar 11 '13 at 11:07

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