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Let $\mathcal{C}$ be a minimal cyclic $A_{\infty}$ category, as considered in, say Kajiura's thesis, or Kontsevich and Soibelman's deformation theory notes, or http://arxiv.org/abs/math/0412149. I.e. it's a minimal $A_{\infty}$ category in the usual sense, with a nondegenerate symmetric pairing $\langle\bullet,\bullet\rangle$ between $\mathrm{Ext}^i(x,y)$ and $\mathrm{Ext}^{n-i}(y,x)$ such that $\langle m_s(\bullet,\ldots,\bullet),\bullet\rangle$ is cyclically invariant (up to the usual sign horror) for every $s$, where the $m_s$ are the (higher) composition operations in the category $\mathcal{C}$. The number $n$ here is fixed throughout.

Now say that $x$ and $y$ are isomorphic objects of $\mathcal{C}$, in the sense that there are morphisms $f:x\rightarrow y$ and $g:y\rightarrow x$, such that $m_2(f,g)=\mathrm{id}_y$ and $m_2(g,f)=\mathrm{id}_x$.

My question is: is there a quick proof that $\mathrm{End}_{\mathcal{C}}(x)$ and $\mathrm{End}_{\mathcal{C}}(y)$ are isomorphic as cyclic $A_{\infty}$ algebras? E.g. we want a morphism $F$ of $A_{\infty}$ algebras such that $F_1$ pulls back the pairing $\langle\bullet,\bullet\rangle$ and there are equalities $\sum_{i+j=t}\langle F_i,F_j\rangle=0$ for all $t\geq 3$. While this is hopefully true, I can't find a reference anywhere, and unless f and g are strict, in the sense that they vanish for all higher compositions, I don't see a quick proof.

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    $\begingroup$ As an update: it seems that it is possible to prove this statement by first running some homological perturbation theory argument, basically the same as the approach to strictifying units found in Lefevre-Hasegawa's thesis, but correcting with Hamiltonian vector flows instead of arbitrary isomorphisms of formal pointed dg manifolds, since we're in the cyclic setting. The outcome of this argument is that the action of the two object category containing just x and y can be replaced with one for which $\textrm{id}_x$, $\textrm{id}_y$, $f$ and $g$ are all strict, then the result is easy. $\endgroup$
    – Ben Davison
    Nov 21, 2012 at 0:22
  • $\begingroup$ E.g. just take $F_1(a)=m_2(f,m_2(a,g))$ and $F_i=0$ for all $i\geq 2$. This isn't a quick proof (the HPT is a bit involved), but I guess it's the best we can do. Although I'd be happy to be proved wrong. $\endgroup$
    – Ben Davison
    Nov 21, 2012 at 0:26

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