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 = $\{\left(x,y\right)\in \mathbb R^{n+k} \mid x \in U, y=f(x) \}$
Can anyone help me please?
 A: 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.
A: 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 $U \subset \mathbb R^n$. 
