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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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2 Answers 2

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

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

and

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

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
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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
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CW complexes are the opposite of "pathological". –  Aaron Mazel-Gee Mar 10 '13 at 19:12
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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
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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 ncatlab.org/nlab/show/Strom+model+structure), together with the fact that all objects are fibrant in the Strøm model structure. –  Ricardo Andrade Mar 11 '13 at 1:34
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@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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