Mapping Out(F_n) to the mapping class group Let $\mathrm{Out}(F_g)$ denote the automorphism group of a free group, and $\mathrm{Mod}_g$ the mapping class group of a closed oriented genus $g$ surface. Is there a map, as indicated with the dashed arrow below, making the following diagram commute?
$$
\begin{array}{ccc}
    \mathrm{Out}(F_g) & \dashrightarrow & \mathrm{Mod}_g \\
    \downarrow & & \downarrow \\
    \mathrm{GL}(g,\mathbf Z) & \to & \mathrm{Sp}(2g,\mathbf Z) \\
\end{array}
$$
The lower horizontal map is induced by the functor $V \mapsto V \oplus V^\ast$ taking a vector space of dimension $g$ to a $2g$-dimensional symplectic vector space.
I do have a reason for asking but it's too long of a story to write here.
 A: As pointed out by YCor, $\mathrm{Out}(F_g)$ is not linear (for $g \geq 3$).  Also, the linearity of $\mathrm{Mod}(S_g)$ is unknown.  So the existence of such an embedding would solve a long-standing open question.  I suspect that there is no such embedding.
However, perhaps you would be interested in a substitute.  Let $V_g$ be the three-dimensional handlebody of genus $g$.   Note that $\partial V_g = S_g$.  Let $\mathrm{Mod}(V_g)$ be the resulting mapping class group.  Restricting to the boundary gives a monomorphism $r \colon \mathrm{Mod}(V_g) \to \mathrm{Mod}(S_g)$.  On the other hand, mapping classes act (via outer automorphism) on the space's fundamental group.  So we have a epimorphism (as it turns out) $f \colon \mathrm{Mod}(V_g) \to \mathrm{Out}(\pi_1(V_g)) \cong \mathrm{Out}(F_g)$.  Thus instead of a commuting square there is a pentagon.
A: This works for $g=2$, but it’s a very special case. $Out(F_2)\cong GL_2(
\mathbb{Z}) \cong Mod_1 \cong Mod_{1,1}$, the mapping class group of a pointed torus. This is realized by the linear action of $GL_2(\mathbb{Z})$ on $T^2=\mathbb{R}^2/\mathbb{Z}^2$ fixing the origin. Taking the oriented blowup of the action at the origin (blowup by rays), one obtains an action of $GL_2(\mathbb{Z})$ on the surface $\Sigma_{1,1}$, a genus 1 surface with one boundary component. Then $GL_2(\mathbb{Z})$ acts on $\Sigma_{1,1}\times [-1,1]$, which is homeomorphic to a genus 2 handlebody and boundary homeomorphic to the double of $\Sigma_{1,1}$ along its boundary, ie $\Sigma_2$ the closed connected orientable surface of genus 2. The first homology splits as a direct sum into $H_1(\Sigma_{1,1} \times \{1\})$ and $H_1(\Sigma_{1,1}\times \{-1\})$, in such a way that the action of $GL_2(\mathbb{Z})$ acts by the dual action on the second factor since the identification by the product with $[-1,1]$ reverses orientation. Hence this gives the sort of homomorphism you seek in this case.
