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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.)

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    $\begingroup$ I don't understand your notation: how are you equating an $n$-tuple $(x^1, \ldots, x^n)$ with an $(n+1)$-tuple $(y^1, \ldots, y^n, f(y^1, \ldots, y^n))$? $\endgroup$ – Rob Arthan Feb 5 '17 at 14:48
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    $\begingroup$ 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. $\endgroup$ – Thomas Feb 5 '17 at 14:51
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    $\begingroup$ 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.) $\endgroup$ – Rob Arthan Feb 5 '17 at 14:56
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    $\begingroup$ 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. $\endgroup$ – Moishe Kohan Feb 5 '17 at 19:24
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    $\begingroup$ 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 $\endgroup$ – Ronald Feb 24 '17 at 9:17
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As was pointed out in the comments, this question only makes sense with reference to an explicit choice of embedding of the manifold in Euclidean space. For second countable manifolds, such an embedding always exists by the Whitney embedding theorem.

Anyway, referring to manifolds explicitly embedded in Euclidean space, it is true that they are all locally the graphs of smooth functions, as a corollary of the implicit function theorem.

The best reference for this fact I could find is Section 4.3 of Krantz, Parks's Implicit Function Theorem. See also this answer on MathOverflow.

This is also addressed on p. 196, Section III.4, of Edwards's Advanced Calculus of Several Variables:

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 $$P = \{ x \in \mathbb{R}^n: (x_{i_1}, \dots, x_{i_k}) \in U \quad \text{and} \quad (x_{i_{k+1}}, \dots, x_{i_n}) = h(x_{i_1}, \dots, x_{i_k}) \}. $$

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