!newcal A
!newcal B
!newop Th
!newop id

\renewcommand{\o}{\circ}
\newcommand{\defn}[1]{{\bf #1}}

:::\newcommand{\terpAr}{\ar@{~>}}
:::\newcommand{\biterpAr}[2]{\terpAr@<0.5ex>[#1]^{#2}}
\newcommand{\terp}[2]{\xymatrix{#1 \terpAr[r] & #2}}

We now address the ``Borel-Tits phenomenon'' associated with 
bi-interpretations which require parameters.

Following \cite{AZ}, but allowing parameters, we say interpretations $f,g : 
\terp{\A}{\B}$ are /homotopic/, written $f \sim g$, if $\{ (f(x),g(x)) : x (- 
\A \}$ is a $\B$-definable set; in other words, the identity map on $\A$ 
induces a $\B$-definable isomorphism between the copies of $\A$ in $\B$ given 
by $f$ and $g$ respectively.

Definition:
  Say a complete theory $T$ has the \defn{unique auto-interpretation property} 
  if whenever $\B,\B' |- T$ and $f : \terp{B'}{\B}$ is an interpretation,
  then $f$ is homotopic to an isomorphism $\theta : \B' --> \B$.
  In other words, $\B'$ is $\B$-definably isomorphic to $\B$.
  .

Theorem:
  Suppose $\sigma : \A_1 --> \A_2$ is an isomorphism,
  and $(f,g)$ is a bi-interpretation of $\A_1$ with $\B_1$,
  and $\alpha : \terp{\A_2}{\B_2}$ is an interpretation,
  and $\Th(\B_1) = \Th(\B_2) =: T_B$, and $T_B$ has the unique 
  auto-interpretation property.

  \[ \xymatrix{
  \A_1 \biterpAr{d}{f} \ar[r]^\sigma & \A_2 \terpAr[d]^{\alpha} \\\
  \B_1 \biterpAr{u}{g} & \B_2 \\\
  } \]

  Then there exist an interpretation $\alpha' \sim \alpha$ and an isomorphism 
  $\sigma' : \B_1 -~=-> \B_2$ such that $\sigma$ is induced from $\sigma'$ 
  according to $f$ and $\alpha'$, i.e. the following diagram of 
  interpretations commutes.
  \[ \xymatrix{
  \A_1 \ar[d]^f \ar[r]^\sigma & \A_2 \ar[d]^{\alpha'} \\\
  \B_1 \ar[r]^{\sigma'} & \B_2 \\\
  } \]
  .
Proof:
  We write composition as concatenation. Then $\alpha \sigma g$ is an 
  interpretation of $\B_1$ in $\B_2$, so by the unique auto-interpretation 
  property there is an isomorphism $\sigma'$ such that $\alpha \sigma g \sim 
  \sigma'$.
  Now $gf$ is definable, i.e. $\id \sim gf$, so $\alpha \sigma \sim \alpha 
  \sigma g f \sim \sigma' f$, so $\alpha \sim \sigma' f \sigma^{-1} =: 
  \alpha'$. Then $\alpha'$ is as required.
  .

\begin{thebibliography}{AZ}
  \bibitem[AZ]{AZ}Ahlbrandt and Ziegler, quasi-finitely axiomatisable totally 
    categorical structures
\end{thebibliography}
