I really wish that, if $\pi:(M,\mathrm{g})\twoheadrightarrow(N,\mathrm{h})$ is a Riemannian covering, then $\mathfrak{i}(N,\mathrm{h})\leq\mathfrak{i}(M,\mathrm{g})$, where $\mathfrak{i}(M,\mathrm{g})$ denotes the Lie algebra of the isometry group of $(M,\mathrm{g})$. I concocted a "proof," but I don't believe it is correct. Could someone give this a sanity check?
Proof: Consider the mapping $$\pi_*^{-1}:\mathfrak{i}(N,\mathrm{h})\to\Gamma(TM),~X\mapsto(p\mapsto\pi^{-1}_*(X_{\pi(p)})\cap T_pM).$$
Already, I'm not sure that this is kosher. This looks too disgusting to be right, but I'm almost positive that this map is well-defined.
Suppose $U\subseteq M$ is open and that $\pi(U)$ is diffeomorphic to $U$. Then, on $U$, $$\mathcal{L}_{\pi_*^{-1}X}\mathrm{g}=\mathcal{L}_{\pi_*^{-1}X}\pi^*\mathrm{h}=\frac{d}{dt}|_0\exp(t\pi_*^{-1}(X))^*\pi^*\mathrm{h}=\frac{d}{dt}|_0(\pi|_U^{-1}\circ\exp(tX)\circ\pi)^*\pi^*\mathrm{h}\\=\frac{d}{dt}|_0(\exp(tX)\circ\pi)^*\mathrm{h}=0.$$
Thus, $\pi_*^{-1}(\mathfrak{i}(N,\mathrm{h}))\subseteq\mathfrak{i}(M,\mathrm{g})$.
This, too, makes me uncomfortable. The above demonstrates that $\pi_*^{-1}$ takes Killing vectors from $N$ to Killing vectors in $M$. The next part proves it is a monomorphism of Lie algebras.
If $\pi_*^{-1}(X)=\pi_*^{-1}(Y)$, then we have $X_{\pi(p)}=Y_{\pi(p)}$ for all $p\in M$. Since $\pi$ is a surjection, it follows that $X=Y$.
This seems right to me. The fact that $\pi_*^{-1}$ is a Lie algebra homomorphism is simply an exercise in computation.
Can anyone help me see a flaw in my proof?