This question is inspired by [this question](https://mathoverflow.net/questions/46597/why-does-the-grothendieck-group-k-0r-of-a-ring-not-depend-on-our-choice-of-us) about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; but I do not know.


Let $G$ be a discrete group, $G^{op}$ its opposite group (i.e. the one with reversed multiplication). Let $H \subset G$ be a subgroup, such that $H=[H,H]=[G,G]$. Note that $H^{op} \subset G^{op}$ has the same properties. I denote the Quillen Plus-Construction of a space with fundamental group $G$ with respect to $H$ by $X \mapsto X^+\_H$



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> **Question:** Is there a homotopy equivalence between $BG\_H^+$ and $B(G^{op})\_{H^{op}}^+$, such that the induced map on $\pi\_1$ is induced by identity $id: G \to G^{op}$.
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Note that there is clearly a homotopy equivalence between $BG\_H^+$ and $B(G^{op})\_{H^{op}}^+$ such that the induced map on $\pi\_1$ is induced by the inverse $inv: G \to G^{op}$; but that is not the one I am looking for.


However, this shows that one could also ask:



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> **Question:** Is there a homotopy equivalence between $BG\_H^+$ and itself, such that the induced map on the abelian group $\pi\_1(BG\_H^+) = G/H$ is the inversion.
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Given the motivation, any good answer in the case $G=GL\_{\infty}(R)$ (for some ring) would be interesting too.


**EDIT:** Johannes Ebert suggested in a comment a strategy to give a negative answer to Question 2 in general. The Kan-Thurston construction gives a way of obtaining every (finite?) cell complex $X$ as the plus-construction on $BG$ with respect to *some* perfect subgroup $H$. It seems that the only missing piece is a finite cell-complex $X$ with $\pi\_1$ abelian and *no* $\pi\_1$-inversion-inducing self-homotopy equivalence. Then, $X = BG^+\_H$ for some group $G$ and some perfect subgroup $H \subset G$. However, since $\pi\_1 = G/H$ is abelian, we see that $[G,G] \subset H$ and hence, $[H,H] = [G,G]=H$ as required. 



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> **Question:** Can anybody give an example of such a space?
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How commutative is Quillen's Plus-Construction?
To answer the final question, here is how you can construct a space with no self-equivalence inducing negation on (abelian) $\pi\_1$.


If $X \to Y$ is a homotopy equivalence, then for any basepoint it induces isomorphisms $\pi\_n(X) \to \pi\_n(Y)$ commuting with the action of $\pi\_1$. In particular, if we can construct abelian groups $A$ and $B$ with an action of $A$ on $B$ such that there is no isomorphism $\phi: B \to B$ such that ${}^a \phi(b) = \phi({}^{(-a)}b)$, then we can use these to construct the desired space; let $X$ be the homotopy orbit space $K(B,2) \times\_A EA$, which has $\pi\_1 = A$ and $\pi\_2 = B$ with the given action.


If $B$ is cyclic, all endomorphisms commute and so it suffices to construct an action so that ${}^a b = {}^{(-a)}b$ does not hold for all $a, b$.


"Minimal" examples include the action of $\mathbb{Z}/4$ on $\mathbb{Z}/5$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 3x$) and the action of $\mathbb{Z}/3$ on $\mathbb{Z}/7$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 4x$).


