# How do I prove that a graph is a topological manifold

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

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

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.

• it's true, I want to prove just that M is a topological manifold – user42912 Oct 3 '12 at 4:04
• 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 $U \subset \mathbb R^n$.

• 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
• 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
• 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
• 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