Torelli theorem for stable vector bundle Let $X, X^{\prime}\colon$ smooth projective curve on $\mathbb{C}$ (genus $\geq 3$),
$M(r,d)\colon$ coarse moduli of stable vector bundles with rank $r\geq2$, and degree $d$ , and
$M(r,\xi)\colon$ coarse moduli of stable vector bundles which have fixed determinant $\xi$ with rank $r$.
In other words, define
\begin{eqnarray}
\operatorname{det}\colon &M(r,d)&\rightarrow &\operatorname{Jac}^{d}(X)\\
&E&\mapsto&\wedge^{r} E
\end{eqnarray}
then,
\begin{equation}
M(r,\xi)=\operatorname{det}^{-1}(\xi)
\end{equation}

There is a famous fact,
$M(r,\xi)_{X}\cong M(r,\xi^{\prime})_{X^{\prime}}$ then $X\cong X^{\prime}$.
I want to know whether $M(r,d)_{X}\cong M(r,d)_{X^{\prime}}$ implies $X\cong X^{\prime}$.
I think it is possible to prove that by almost the same proof.
But if not, why I must consider fixed determinant ? 
Thanks in advance.
 A: As dhy said, this follows if we know that the fibers $M(r,\xi)$ are rational. In fact, it suffices to know that they are unirational - we can deduce that their image in any map to an abelian variety is a point, hence any map from $M(r,d)$ to an abelian variety factors through the determinant map, and thus the Albanese of $M(r,d)$ is $\operatorname{Jac}^d$.
Their unirationality is not so hard to prove.
Take $L$ ample enough that $V \otimes L$ is globally generated for every stable vector bundle of rank $r$ and determinant $\xi$. (It suffices to have $H^1( X, V \otimes L (-P)) =0$ for all points $P$, i.e. it suffices to have $H^0(X, V^\vee \otimes K_X \otimes L^{\vee} (P))$, so it suffices to have $\deg L > 2g-1 + \frac{ \deg xi}{r}$.)
Then among maps  $(L^{-1})^{n-1} \to V$, those which have rank $< n-1$ at a point $P$ form a codimension $2$ subset, so those which have rank $<n-1$ at any point form a codimension $1$ subset, and thus there exists a map $(L^{-1})^{n-1} \to V$ with full rank at every point. Hence the quotient is a line bundle, which because $\det V = \xi$, must be $ \xi \otimes L^{\otimes (n-1)}$. So $V$ is an extension of $\xi \otimes L^{\otimes (n-1)}$ by $(L^{-1})^{n-1}$.
Now there is an open subset of $\operatorname{Ext}^1 ( \xi \otimes L^{\otimes (n-1)}, (L^{-1})^{n-1} )$ parameterizing stable vector bundles, which maps to $M(r,\xi)$. By the above argument this map is surjective, so $M(r,\xi)$ is unirational.
