Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What is "aberto"? –  Paul Oct 3 '12 at 2:37
2  
Portugese for open. –  KReiser Oct 3 '12 at 4:10
    
@Paul sorry, my fault –  user42912 Oct 3 '12 at 14:13

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
it's true, I want to prove just that M is a topological manifold –  user42912 Oct 3 '12 at 4:04
1  
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 $\mathbb R^{n+k}$. –  user42912 Oct 3 '12 at 14:33
    
@user42912 is $\Psi$ continuous as constructed? Is it injective? –  James S. Cook Oct 4 '12 at 6:23
    
yes it's continuous, because its components functions are continuous and it's injective. Is it a homeomorphism? –  user42912 Oct 16 '12 at 10:00
    
@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. –  James S. Cook Oct 18 '12 at 6:55

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 $U \subset \mathbb R^n$.

share|improve this answer
2  
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). –  commenter Oct 3 '12 at 15:40
    
@commenter Hah, true. Let me think about this and fix it, I think the idea I had is right. –  Rudy the Reindeer Oct 3 '12 at 16:12
2  
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 $M$. –  commenter Oct 3 '12 at 19:48
2  
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 $\Psi$. –  commenter Oct 10 '12 at 10:52
1  
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... –  commenter Oct 12 '12 at 15:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.