The question is in the title : are there spaces X such that the adjoint of the identity on the loop space $\Omega X$, i.e. $\Sigma\Omega X \to X$, is a homotopy equivalence ?
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If $X$ is such a space (a CW complex, say), then it must be a suspension $\Sigma Y$ (as it is the suspension of $y :=\Omega X$). The James splitting gives $$\Sigma \Omega \Sigma Y \simeq \bigvee_{n=1}^\infty \Sigma Y^{\wedge n},$$ the wedge of suspensions of smash products, so under your assumption $\Sigma \Omega \Sigma Y \overset{\sim}\to \Sigma Y$ it follows that $\Sigma Y^{\wedge n} \simeq *$ for all $n > 1$. In particular, by the Kunneth theorem $Y$ must have the homology of a point, and in particular be path-connected. But then $X=\Sigma Y$ must be simply-connected, and also have the homology of a point: therefore it is contractible.
answered Sep 22, 2015 at 9:35
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$\begingroup$ There's a step I don't see. If you take $Y=M(\mathbb Q/\mathbb Z,n)$, $n\geq 2$, $Y^{\wedge 2}$ is contractible since $\mathbb Q/\mathbb Z\otimes \mathbb Q/\mathbb Z=0=Tor_1^{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)$, isn't it? $\endgroup$ Commented Sep 22, 2015 at 9:52
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1$\begingroup$ The $Tor_1$ isn't zero: There's an exact sequence $Tor_1(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \rightarrow Tor_1(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \rightarrow \mathbb{Z}\otimes \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q}\otimes \mathbb{Q}/\mathbb{Z}$. The right- and leftmost terms are 0, so the Tor is just $\mathbb{Q}/\mathbb{Z}$. $\endgroup$ Commented Sep 22, 2015 at 10:03
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$\begingroup$ @AchimKrause OMG, I took an injective module for a flat one! $\endgroup$ Commented Sep 22, 2015 at 10:11
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5$\begingroup$ In general, a map that induces isos on homology with field cofficients induces isos on integral homology. A quick rundown: For $A\rightarrow B\rightarrow C$ a short exact sequence of abelian groups, if it works for two of these coefficients, it works for the third - that's the five-lemma. Start from $\mathbb{F}_p$ to inductively prove it for all $\mathbb{Z}/p^n$. Then it also follows for their colimit, $\mathbb{Z}/p^{\infty}$. But $\mathbb{Q}/\mathbb{Z}$ decomposes into copies of $\mathbb{Z}/p^{\infty}$, so we're done by $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$. $\endgroup$ Commented Sep 22, 2015 at 10:12
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$\begingroup$ @AchimKrause: Maybe I should know this, but what exactly does the notation $\mathbb{Z}/p^\infty$ mean? $\endgroup$ Commented Sep 23, 2015 at 4:17