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I'm having problems with that:

Prove that M is a topological manifold.

$f:U \to \mathbb R^k$, U $\subset \mathbb R^{n}$ open, f continuous

M = $\{\left(x,y\right)\in \mathbb R^{n+k} \mid x \in U, y=f(x) \}$

Can anyone help me please?

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    What is "aberto"?2012-10-03
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    Portugese for open.2012-10-03
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    @Paul sorry, my fault2012-10-03

2 Answers 2

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Well, $\Psi(x) = (x,f(x))$ provides a patch from $U \subseteq \mathbb{R}^n$ into $\mathbb{R}^{n+k}$. However, given what you say thus far I think I can at most say $M$ is a topological manifold. We need further data about $f$ to say more.

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    it's true, I want to prove just that M is a topological manifold2012-10-03
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    I can't understand why it's a topological manifold. For each point in M we have to find a neighborhood and a homeomorphism between this neighborhood and an open subset of $\mathbb R^{n+k}$.2012-10-03
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    @user42912 is $\Psi$ continuous as constructed? Is it injective?2012-10-04
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    yes it's continuous, because its components functions are continuous and it's injective. Is it a homeomorphism?2012-10-16
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    @user42912 precisely. Note the surjectivity is clear as the codomain $U \times f(U)$ is clearly attained. The injectivity of $\Phi$ is clear from the $x$ in $(x,f(x))$. And continuity can also be seen since $\Phi$ is the cartesian product of continuous maps.2012-10-18
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To show that $M$ is a topological $r$-manifold you would like to show that every point $m$ in $M$ is contained in an open set that is homeomorphic to an open subset of $\mathbb R^r$.

Maybe we should first think about what the dimension $r$ is in this case. Points in $M$ are of the form $(x,f(x)) = (x_1, \dots, x_n, f(x))$. Since $f$ is determined by $x_1, \dots, x_n$ the dimension of $M$ is $n$.

Now let $(x,f(x))$ be a point in $M$. We would like to find an open set containing $(x,f(x))$ and a homeomorphism from the set to an open subset of $\mathbb R^n$. The whole space $M = U \times f(U)$ is of course open and contains $(x,f(x))$. If we can find a homeomorphism from $U \times f(U)$ to an open subset of $\mathbb R^n$ then we're done. As pointed out by commenter in the comments, the map $h: U \to U \times f(U), x \mapsto (x,f(x))$ is continuous and bijective and its inverse $h^{-1}: U \times f(U) \to U, (x,f(x)) \mapsto x$, which is the projection, is also continuous hence $h$ is a homeomorphism between $M$ and $U \subset \mathbb R^n$.

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    Why is $f^{-1}(O) \times O \subset M$? (It can't be true because $M$ is $n$-dimensional and $f^{-1}(O) \times O$ is $n+k$-dimensional).2012-10-03
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    @commenter Hah, true. Let me think about this and fix it, I think the idea I had is right.2012-10-03
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    Use what James said: if $f$ is continuous then $\Psi \colon x \mapsto (x,f(x))$ is a homeo of $U$ onto $M$ because $M \ni (x,y) \mapsto x \in U$ is a continuous inverse of $\Psi$. Thus, $\Psi$ maps open subsets of $U$ to open subsets of $M$.2012-10-03
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    @commenter Sorry for taking so long. If $h^{-1}: U \times f(U) \to U$ is the inverse, how did you prove that it's continuous? If $V \subset U$ is open then $(h^{-1})^{-1}V = h(V) = V \times f(V)$ might not be open.2012-10-10
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    The projection $\pi \colon \mathbb{R}^n \times \mathbb{R}^k \to \mathbb{R}^n$ is certainly continuous. The inverse of $\Psi$ is the restriction of $\pi$ to the graph of $\Psi$.2012-10-10
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    @commenter Oh dear! :,( Thank you. I'll rewrite and then delete it -- I don't think duplicate information is necessary. I'm doing this to practice (also the reason why I posted an answer).2012-10-10
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    Dear @commenter, I edited it. Is it correct now?2012-10-11
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    Looks good. I think you can leave this answer up (no need to delete it), it might be useful for those who didn't follow James's hint...2012-10-12
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    @commenter Thanks loads for reading it! Ok, then I will not delete it.2012-10-12