Claim. The restriction map is always proper, where the target group $G$ is the group of $K$-points of a reductive group over a local field $K$, e.g. $G=GL_N({\mathbb C})$.

Proof. First, some generalities, details for which you can find, for instance, here. Let $X$ be the symmetric space or a locally compact Euclidean building corresponding to $G$. For each representation $\rho$ to $G$ of a group $\Lambda$ with generators $\gamma_1,...,\gamma_k$, define the min-max displacement
$$
d_\rho:=\inf_{x\in X} d_\rho(x), d_\rho(x):=\max_j \rho(\gamma_j)(x).
$$
Let $x_\rho\in X$ denote a point for which $d_\rho(x)-d_\rho\le 1$. It is not hard to check that the number $d_\rho$ depends only on the projection $[\rho]$ of $\rho$ to $M(\Lambda, G)$. Furthermore, a sequence $[\rho_i]$ is precompact in $M(\Lambda, G)$ iff the sequence $(d_{\rho_i})$ is bounded. Suppose that $[\rho_i]$ is not precompact and the sequence $(d_{\rho_i})$ diverges to infinity. Then you take the asymptotic cone $Cone(X)$ of $X$ centered at the points $x_{\rho_i}$ with scaling factors $(d_{\rho_i})^{-1}$. The result is an isometric action $\rho$ of $\Lambda$ on $Cone(X)$ without a common fixed point. The key fact is that $Cone(X)$ is a Euclidean building, by a theorem of Kleiner and Leeb. By the Cartan-Tits theorem, the action $\rho$ has no bounded orbits.

Now, we can prove the claim. Suppose to the contrary, that there exists a sequence of representations $\rho_i$ of $\Gamma$ whose projections to $M=M(\Gamma, G)$ diverge, while their restrictions $\rho'_i$ to $\Gamma_0$ project to a relatively compact sequence in $M(\Gamma_0,G)$. Then the above construction yields an action $\rho$ of $\Gamma$ on $Cone(X)$.

Note that the restricted actions $\rho_i'$ have bounded $d_{\rho'_i}$. There are two cases to consider:

The distance between points $x_{\rho_i'}$ and $x_{\rho_i}$ is $O(d_{\rho_i})$. Then
the sequence $(x_{\rho_i'})$ represents a point $x$ in $Cone(X)$. This point is necessarily fixed by $\Gamma_0$. Thus, since $|\Gamma:\Gamma_0|<\infty$, the $\Gamma$-orbit of $x$ is bounded, which is a contradiction.

$d_{\rho_i}=o(d(x_{\rho_i'},x_{\rho_i}))$. Then geodesic segments between $x_{\rho_i'},x_{\rho_i}$ represent a geodesic ray in $Cone(X)$. The point at infinity $\xi$ represented by this ray is fixed by $\Gamma_0$ and, moreover, every element of $\Gamma_0$ acts as a unipotent isometry of $Cone(X)$. Hence, $\Gamma_0$ has a common fixed point in $X$. Now, the argument is the same as in Case 1. QED.

finite(in the sense of algebraic geometry), i.e. proper with finite fibres. Concretely, given an $n$-dimensional rep'n of $\Gamma_0$, won't there be at most finitely many non-isomorphic ways to extend it to an $n$-dimensional representation of $\Gamma$ if the index $[\Gamma:\Gamma_0]$ is finite? Regards, – Emerton Jul 10 '12 at 15:00