Why is $\{x\in M: F(x)=G(x), DF_x=DG_x\}$ closed?

Question

Let $F$ and $G$ be smooth maps between smooth manifolds $M$ and $N$. Denote their differentials at $x\in M$ by $DF_x$ and $DG_x$. Why is the set $\{x\in M:F(x)=G(x), DF_x=DG_x\}$ closed in $M$?

I know of the result that if $f$ and $g$ are continuous maps between topological spaces $X$ and $Y$ with $Y$ being Hausdorff then $\{x\in X:f(x)=g(x)\}$ is closed in $X$. Unfortunately it doesn't seem like I can directly apply this result since I have $DF_x$ and $DG_x$ which are now maps between tangent spaces.

Answer 1

It suffices to prove that if $f: M \to N$ is smooth, then

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \}$

is closed in $M$ (just consider $f = F - G$ and $N$ embedded in some Euclidean space). Now,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = \{ x \in M : f(x) = 0\} \cap \{ x \in M : Df_x = 0\}.$

Since $f$ is continuous, $f^{-1}(\{ 0\})$ is closed in $M$. Since $f$ is smooth, $Df : TM \to TN$ is also continuous as a map between tangent bundles. So, the map $E : M \to \text{Hom}(TM, TN)$ given by

$E(x) = Df_x, \quad x \in M,$

is also continuous. Thus, $E^{-1}(\{0\})$ is also closed in $M$. Hence,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = f^{-1}(\{0\}) \cap E^{-1}(\{0\})$

is closed, as required.

Answer 2

Take a sequence $x_n\to x$ and note that if $$ F(x_n)=G(x_n)\quad\text{and}\quad DF_{x_n}=DG_{x_n} $$ for all $n$, then $$ F(x)=G(x)\quad\text{and}\quad DF_{x}=DG_{x}. $$