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Let $f$ be a $C^k$ map defined in an open set $U\in \mathbb R^n$, I'm trying to prove this is a manifold. I couldn't find a parametrization. I've tried $\varphi(x_1,\ldots,x_n)=(x_1,\ldots,x_n,f(x_1,\ldots,x_n))$, but I couldn't prove this is an homeomorphism. I need help here.

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    Where do you get stuck in showing that $\phi$ is a homeomorphism? – 2017-02-12
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    @littleO If $f$ is not $1-1$, then $\varphi$ isn't $1-1$ either – 2017-02-12
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    Actually, $\phi$ is $1-1$ even if $f$ is not $1-1$. I bet you could prove that. You could start by supposing that $\phi(x_1,\ldots,x_n) = \phi(y_1,\ldots,y_n)$. – 2017-02-12
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    @littleO yes, you're right, I've just proved. It's easy. To prove the onto part, can I restrict the image of $\varphi$ to be only the graph of $f$? – 2017-02-12
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    I think you can show that $\phi$ is onto by using the definition of the "graph" of a function. Suppose that a point $p$ belongs to the graph of $f$. By the definition of the graph of a function, $p$ has the form $p = \ldots$ – 2017-02-12
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    Yes, I got it. Thank you! – 2017-02-12

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You've got the right parameterization. It's certainly a 1-1 map (why?), and certainly onto (why?), and certainly continuous (why?), and indeed, $C^k$ (why? What's the derivative matrix at $(x_1, \ldots, x_n)$ look like?).

Hint: The only possible remaining part is finding an inverse and proving that the inverse is nice as well. So: here's the inverse. Let $K$ denote the graph of $f$.

$$ p: K \to \Bbb R^n : (x_1, \ldots, x_n, u_{n+1}, \ldots, u_k) \mapsto (x_1, \ldots, x_n). $$

Can you explain why $p$ is continuous, 1-1 (on $K$), and $C^k$?

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    Can I say $\varphi$ is continuous because $\pi_i(x_1,\ldots,x_n)=x_i$ and $f$ are continuous? can I apply the same argument to prove $\varphi$ is a $C^k$ map? – 2017-02-12
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    Yes, $\phi$ is continuous because each component of $\phi$ is continuous. The same kind of argument works for $C^k$. – 2017-02-12
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    John, how would one prove that the graph is an open set? – 2018-11-12
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    Well, it's the preimage of an open set (namely $U$) under a continuous map (namely $p$). Of course, that only makes $K$ open **in $K$**. Then again, that's the only notion that makes sense. In general it's *not* open in $\Bbb R^n$; look at the graph of $f: \Bbb R \to \Bbb R : x \mapsto 0$ in $\Bbb R^2$: the graph is a horizontal line; its complement is the union of two open halfspaces in $\Bbb R^2$, hence open, so the graph is (as a subset of $\Bbb R^2$) closed. It can't be both closed **and** open, because it's not a component of $\Bbb R^2$. Hence it's not open (as a subset of the plane). – 2018-11-12