Is $f^{-1}(1)$ a submanifold of $\Bbb S^2\times \Bbb S^2$?

Question

We consider $f:\Bbb S^2\times\Bbb S^2\to\Bbb R$ as the standard dot product of two points in $\Bbb R^3$. We need to prove that $f^{-1}(1)$ is a regular submanifold of $\Bbb S^2\times\Bbb S^2$.

If $1$ were a regular value, then it would be direct, but it isn't (the jacobian of $f$ is zero in every two points $x$ and $y$ such that $x=y$, and $f(x,x)=1$). Also, the rank of $f$ is not constant (since it can vary from $1$ and $2$).

I don't understand well how to prove it by definition. Could anyone please help me out?

Answer 1

The following result can be found in Lee's Introduction to Smooth Manifolds, Proposition 5.7.

Let $M, N$ be smooth manifold and let $g : M\to N$ be a smooth map. Then the graph of $g$ $$ \Gamma_g =\{ (x, g(x)) \in M \times N : x\in M\}$$ is an embedded submanifold of dimension $m = \dim (M)$.

Setting $M = N = \mathbb S^2$ and $g : \mathbb S^2 \to \mathbb S^2$, $g(x) = x$, then

$$D = \{ (x, x) \in \mathbb S^2 \times \mathbb S^2 : x\in \mathbb S^2\}$$ is an embedded submanifold of $\mathbb S^2 \times \mathbb S^2$. Since $f^{-1}(1) = D$, $f^{-1}(1)$ is an embedded submanifold of $\mathbb S^2 \times \mathbb S^2$.