proving a particular subset of a Riemannian manifold is closed using continuity

Question

I have two Riemannian manifolds, $(M,g)$ and $(\widetilde{M},\widetilde{g})$ and two maps $\varphi, \psi : M \to \widetilde{M}$, which are both local isometries. I am trying to show that the set $A = \{ p \in M \mid d \varphi_p = d \psi_p \}$ is closed in $M$ (where $d \varphi_p$ is the differential at the point $p$). I know that I should be able to use continuity of the global differentials $d \varphi$ and $d \psi$ to prove this, but I keep running into problems when I try to write $A$ as the preimage of a closed set. For example, I tried writing $A = \pi((d \varphi - d \psi)^{-1}(\{ 0_{\widetilde{q}} \mid \widetilde{q} \in \widetilde{M} \}))$, but this will give me that $A = M$ even if I don't have that $d \varphi_p = d \psi_p$ for all $p \in M$. I'm sure there is something obvious that for whatever reason I'm just not seeing... what is it?

Answer 1

For an elementary reasoning you can use the fact that (in finite dimensional Riemanian manifolds), for a set $A$ to be closed, it is sufficient to show that $a_n\rightarrow a$ with $a_n\in A$ implies $a\in A$. This follows from continuity of the differentials.