# Does every manifold have an atlas consisting of graph coordinates?

A pointer to a reference will suffice for an answer.

1. Given an $n$-dimensional smooth manifold $M$, do we have an atlas $\{(U, \phi)\}$ for $M$ such that each $\phi$ has the form: $$\phi(p)=(x^1, \dots, x^n)=(y^1, \dots, y^n, f(y_1, \dots, y^n))$$ for a smooth function $f: \mathbb{R}^n \to \mathbb{R}$? I.e. can the manifold be "divided into pieces", such that each "piece" is diffeomorphic to the graph of a smooth function?

2. Does the analogous statement hold for topological manifolds? (Replacing "smooth manifold" by "topological manifold" and "smooth function" by "continuous function".)

3. If a second countable, Hausdorff space is locally expressible as the graph of a continuous function, then is it a topological manifold?

(Yes, because the identification $(x^1, \dots, x^n) \leftrightarrow (y^1, \dots, y^n, f(y^1, \dots, y^n))$ is a homeomorphism?)

Note: this question is a duplicate of this unanswered question. For smooth manifolds, the answer obviously involves the implicit function theorem.

However, I am not only considering manifolds which are already explicitly embedded or immersed in $\mathbb{R}^n$. (I.e. I am asking about intrinsic geometry of manifolds not extrinsic.) To the best of my understanding, an embedding could be used to construct such an atlas, but a general immersion cannot. If this understanding is correct, then the answer to 1. in the affirmative follows from Whitney's theorem.)

• I don't understand your notation: how are you equating an $n$-tuple $(x^1, \ldots, x^n)$ with an $(n+1)$-tuple $(y^1, \ldots, y^n, f(y^1, \ldots, y^n))$? – Rob Arthan Feb 5 '17 at 14:48
• The answer to 1 follows from the definition of manifolds. As long as you are not trying to respect additional structures like a Riemannian metric, just take any chart $\phi=(x_1, \ldots, x_n)$ and let $f=0$ (as @RobArthan pointed out your notation is not consistent. Either you have $n$ components or $n+1$) Edit: things become difficult and require additional reasoning or even assumptions if you either ask questions about global results (i.e. can I embed all of $M$ in a certain way) or add additional structure, like a metric. – Thomas Feb 5 '17 at 14:51
• The question you claim to be duplicating is about an exercise where (although the OP didn't say so) the manifold is embedded in $\Bbb{R}^N$ for some $N$. I don't see how you can have a sensible result of the kind you want without reference to such an embedding. (As @Thomas points out, it is easy to find a diffeomorphism between any open subset of $\Bbb{R}^n$ and the graph of a function.) – Rob Arthan Feb 5 '17 at 14:56
• For a topological submanifold of $R^N$ the answer is negative. Just think about the Koch snowflake. Part 3 of your question is meaningless as written. – Moishe Kohan Feb 5 '17 at 19:24
• If you take $\mathbb R^2$ and$(x,y)\mapsto (x,f(x))$ then then $dim T_{(x,y)}\mathbb R^2=1$! So I think it works only for one dimensional manifolds – Ronald Feb 24 '17 at 9:17

Recall that a $k$-dimensional manifold in $\mathbb{R}^n$ is a set $M$ that looks locally like the graph of a mapping from $\mathbb{R}^k$ to $\mathbb{R}^{n-k}$. That is, every point of $M$ lies in an open subset $V$ of $\mathbb{R}^n$ such that $P = V \cap M$ is a $k$-dimensional patch. Recall that this means there exists a permutation $x_{i_1}, \dots, x_{i_n}$ of $x_1, \dots, x_n$, and a differentiable mapping $h: U \to \mathbb{R}^{n-k}$ defined on an open set $U \subset \mathbb{R}^k$, such that $$P = \{ x \in \mathbb{R}^n: (x_{i_1}, \dots, x_{i_k}) \in U \quad \text{and} \quad (x_{i_{k+1}}, \dots, x_{i_n}) = h(x_{i_1}, \dots, x_{i_k}) \}.$$