Show that graph $ \Gamma(f) = \{x,f(x) : x \in \mathbb{R^n}\}$ of $f$ is a regular submanifold

Question

Let $f : \mathbb{R^n} \rightarrow \mathbb{R^m}$ be a smooth map. Then show that the graph $\Gamma(f) = \{ (x,f(x) ) : x \in \mathbb{R^n} \}$ of $f$ is a regular submanifold of $ \mathbb{R^n} \times \mathbb{R^m}$ of dimension $n$.

I need to use the Regular level set theorem here but I am not able to apply it here. I suppose if $x$ is a regular point in $\mathbb{R^n}$ then $f(x)$ has to be a regular value in $\mathbb{R^m}$. But I am completely stuck here. Any help!

Answer 1

Consider $h:R^n\times R^n\rightarrow R^n$ defined by $h(x,y)=f(x)-y$, $0$ is a regular value since $dh_{(x,y)}(0,v)=-v$ and $\Gamma(f)=h^{-1}(0)$.