Is the subgroup of homotopically trivial isometries a closed subgroup of the isometry group? Let $(M,g)$ be a connected Riemannian manifold. Then according to the Steenrod-Myers-Theorem, the isometry group $\text{Isom}(M,g)$ of $(M,g)$ is a compact lie group with the compact-open topology.
Is the subgroup $G$ of isometries which are homotopically trivial (i.e. homotopic to the identity) a closed subgroup of $\text{Isom}(M,g)$?
Background: We can choose a basepoint $x \in M$ and consider the group homomorphism
$$
\varphi \colon \text{Isom}(M,g) \to \text{Out}(\pi_1(M,x))
$$
which maps an isometry $f$ to the class of the automorphism
$$
[\gamma] \mapsto [P \;*\; (f \circ \gamma) \;*\; \overline{P}]
$$
of $\pi_1(M,x)$, where $P$ is any path in $M$ from $x$ to $f(x)$ and $*$ is concatenation of paths.
Now if $M$ is aspherical, then $G$ is the kernel of $\varphi$ and its plausible that with the right choice of a topology on the outer automorphism group, $\varphi$ is a continuous map and thus $G$ is closed.

Let $I$ denote the isometry group of $(M,g)$. Being a Lie group, it is locally path connected. It follows that the subgroup $G< I$ you are interested in, is open. Now, it is a nice exercise to work out is that each open subgroup of an arbitrary topological group is also closed. 
