# Question on the implicit function theorem and dimension.

Let $f : \mathbb{R}^{n+m} → \mathbb{R}^{m}$ be a continuously differentiable function. Let $(x, y) = (x_1, ..., x_n, y_1, ..., y_m)$. Suppose $f$ satisfies the necessary condition for the implicit function theorem. Then we obtain open set $U \in \mathbb{R}^n$ and $g: \mathbb{R}^n → \mathbb{R}^m$ whose graph $(x, g(x))$ is precisely the set of all $(x, y)$ such that $f(x, y) = 0$, i.e. $$\{ ( x , g ( x ) ) ∣ x ∈ U \} = \{ ( x , y ) ∈ U × V ∣ f ( x , y ) = 0 \}.$$

I would like to think that this set $\{ ( x , g ( x ) ) ∣ x ∈ U \}$ has 'dimension $n$'. I was wondering is there a way to state this more rigorously?

$\{(x,g(x)|x\in U\}$ is a $n$-dimensional manifold