Proving that $\Delta(M \times M)$ is a submanifold of $M \times M$

Question

I am struggling to prove that

$\Delta(M \times M) = \{(x,x) : x \in M\}$ is a submanifold of $M \times M$.

A manifold M is a submanifold of N if there is an inclusion map $i:M \rightarrow N$ such that:

1. $i$ is smooth

2. $Di_x$ is injective for each $x \in M$

3. The manifold topology of $M$ is the induced topology from $N$.

I also know the following theorem:

Let $F:M \rightarrow N$ be a smooth map and $c \in N$ be such that at each point $a \in F^{-1}(c)$ the derivative $DF_a$ is surjective.

Then, $F^{-1}(c)$ is a smooth manifold of dimension $dim(M)-dim(N)$.

In the course of the proof, we see that the manifold structure on $F^{-1}(c)$ satisfies the conditions of the definition of submanifolds.

Problem:

We want to avoid the verifications of the definition of submanifold.

Applying the theorem, we can take $F:M \times M \rightarrow M$ defined by $(x,y) \mapsto y-x$, and so $\Delta(M \times M) = F^{-1}(0)$.

But $y-x$ in general may not belong to $M$.

What do I need to do to prove what I want to prove? Hints? Thanks

Answer 1

I'm actually going to expand my comment. Refer to Lee's Smooth Manifolds proposition 5.4, which describes that, if $M,N$ are smooth manifolds, $U\subset M$ is open and $f:U\to N$ is smooth, then $\Gamma(f)=\{(x,f(x))\in M\times N\}$ is a smoothly embedded submanifold of $M\times N$ and $\dim \Gamma(f)=\dim M$.

Now choose $M=N$ and $U=M$ and consider $id: M\to M$. You just need to verify that this is a smooth map, and then $\Delta(M)=\Gamma(id)$ is a smoothly embedded submanifold. Actually, we're even better since $\Delta(M)\subset M\times M$ is closed since $M$ is Hausdorff, so $\Delta(M)$ is a properly embedded submanifold.

Answer 2

Hint: Consider the map $M \to M \times M$ given by $x \mapsto (x,x)$.