I am asked to prove the following:
"Show that if $f$ is differentiable, then the graph of $f$ is an $n$-dimensional submanifold of $\mathbb{R}^{n+m}$. Does the converse hold?"
I have proved the statement. Now I am faced with the converse. My first thought was: "No, the converse does not hold." And to demonstrate a counterexample, I should pick a continuous function with a cusp like $y = |x|$. But while $|x|$ is a manifold, my understanding is that it is not a submanifold of $\mathbb{R}^2$. So now I am not sure. On the other hand, I have found many proofs of the forward direction only; I assume if the converse were also true then these proofs would be iff-style proofs.
If the converse is false, any hints about a counterexample would be much appreciated! I can work out the details.
(Note, the definition of submanifold we are using in this class is as follows: A subset $M^n \subseteq \mathbb{R}^{n+k}$ is an $n$-dimensional submanifold of $\mathbb{R}^{n+k}$ of class $C^p$ if, for all $x \in M^n$, there exists a neighborhood $U$ of $x$ and a $C^p$ submersion $f: U \rightarrow \mathbb{R}^k$ and $U \cap M = f^{-1}(0)$.)
An example is already suggested in the comments: consider the graph
$$S= \{ (x, x^{1/3}) \in \mathbb R^2: \ x\in \mathbb R\}.$$
First of all, $S$ is a $1$-submanifold in $\mathbb R^2$ since
$$S= \{ (y^3, y) \in \mathbb R^2: \ y\in \mathbb R\}$$ and $y\mapsto y^3$ is smooth. However $x\mapsto x^{1/3}$ is not smooth, so the converse does not hold.