I got a problem
Let $f : \mathbb{R}^m \to \mathbb{R}$ be a $C^k$ function between two Euclidean spaces. And let a set $A$ be a graph of $f$. Show that $A$ is a $C^k$ manifold.
And I solved it.
Could you check if it’s right?
I think it must be wrong, because, in my proof, I didn’t use the condition that $f$ is $C^k$.
Step 1
I showed there is a homeomorphism from $A$ to $\mathbb{R}^m$.
I used the following map, $\varphi: \mathbb{R}^m \to A$ defined by $$\varphi(x_1,...,x_m)=(x_1,...,x_m,f(x_1,...x_m)).$$
It is a bijective map since $f$ is a function. Also we can show that it is continuous.(By using inverse image of open set is open)
And its inverse map is also continuous(just by using the composite an inclusion map and a projection map)
Step 2
Thus $(A,\varphi^{-1})$ is a global chart and thus consists of an atlas. So it can be extended to a maximal atlas.
So, the graph of $f$ is manifold. Actually, it is a smooth manifold. Thus it is $C^k$ manifold.
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I don’t know where I should use $C^k$ condition here.
Thank you!
There's a subtlety in this situation that might not be obvious at first. What you proved (correctly) is that the graph is a topological manifold, and that it has a smooth structure. That's true as long as $f$ is merely continuous.
Sometimes, though, when people ask whether a certain subset of Euclidean space is a $C^k$ manifold, what they really have in mind is whether it is a $\boldsymbol C^k$ submanifold. To show that, you'd need $f$ to be at least of class $C^k$.