Graph of a smooth function on a smooth manifold is again a smooth submanifold Good evening!
If $M$ is a smooth Banach-manifold and $f\in C^\infty(M,M)$ smooth, then it is widely known that the graph $\Gamma(f) = \{(x,f(x)) : x\in M\}$  of $f$ is a smooth submanifold of $M\times M$.
I would like to ask whether or not my proof of this statement is correct:
Take the map $\Phi : M\longrightarrow M\times M, x\longmapsto (x,f(x))$. Then the inverse is given by the projection $\Phi^{-1}: \Gamma(f) \longrightarrow M ,(x,y)\longmapsto x$. Since $f$ was assumed to be smooth, $\Phi$ is a $C^\infty$-diffeomorphism, and so the $C^\infty$-structure of $M$ can be transported to $\Gamma(f)$, making it into a smooth Banach-manifold in its own right.
What I am a bit unsure about is how this induced manifold structure of $\Gamma(f)$ relates to the manifold structure of $M\times M$, i.e. whether or not $\Gamma(f)$ actually becomes a submanifold of $M\times M$ this way. But I think this should be the case, since $M\times M$ inherits its structure from $M$.
Is this line of thought correct?
 A: No, your proof is not right because you didn't link it to the manifold structure of the product, all you showed is there is a smooth structure on the graph for which your $\Phi$ is a diffeomorphism. This is actually very obvious: given a type of mathematical object $M$ (group/ring/field/vector space/inner-product space/Banach space/smooth manifold etc), any set $S$, and a bijection $\Phi:M\to S$, we can always 'transport' the structure from $M$ to $S$ such that $\Phi$ becomes an isomorphism in the appropriate sense.
You can actually generalize the statement. If $f:M\to N$ is a smooth map between smooth manifolds, then its graph, $\Gamma(f)$, is an embedded submanifold of $M\times N$ (equipped with the product manifold structure). You can also weaken the hypothesis to $C^r$ manifolds and $C^r$ maps, and the proof is oblivious to whether you consider finite-dimensional manifolds or Banach manifolds.
To prove this, take any point $(p,f(p))\in \Gamma(f)$. Fix a chart $(U,\alpha)$ in $M$ around the point $p$ and a chart $(V,\beta)$ around the point $f(p)$ in $N$, and also assume that $f$ maps $U$ into $V$ (we can always arrange for this since $f$ is continuous so we can replace $U$ by the smaller open set $U\cap f^{-1}(V)$ if necessary). This gives rise to a 'product chart' $(U\times V,\alpha\times \beta)$, in the obvious way $\alpha\times \beta:U\times V\to \alpha[U]\times\beta[V]$, $(x,y)\mapsto (\alpha(x),\beta(y))$. Now, let us construct a new chart on the product manifold, $(U\times V, \psi_f)$, where the map $\psi_f$ is defined as
\begin{align}
\psi_f(x,y):=(\alpha(x),\beta(y)-\beta(f(x)))
\end{align}
I leave it to you to justify why $(U\times V,\psi_f)$ actually belongs to the maximal atlas on the product (i.e you must show why it is compatible with all other product charts). This is easy to do by-hand because inverting things is very trivial here. Now, the chart $(U\times V,\psi_f)$, when restricted to $\Gamma(f)$ is such that is maps $\Gamma(f)\cap (U\times V)$ onto $\psi_f[U\times V]\cap \alpha[U]\times \{0\}$, i.e $\psi_f$ flattens out the graph in the obvious way, by subtracting the height of the graph at every point. Thus, I've shown that for each point in the graph, there is a chart belonging to the maximal atlas of the product $M\times N$, such that this chart satisfies the 'slice-condition' in (one of the equivalent) definition of an embedded submanifold.
Here, I gave the argument explicitly (leaving minor details for you to fill in) because it is possible to do this without any heavy lifting. There is a more general  theorem, as stated and proved here, relying on the inverse/implicit function theorem. Note that in the link, I stated it for finite-dimensional manifolds, but it generalizes easily to the Banach setting (perhaps one needs to add some conditions regarding existence of closed complementary subspaces), because the inverse function theorem holds in Banach spaces (because one of the ingredients in proving the IFT is Banach's fixed point theorem which holds in complete metric spaces... e.g Banach spaces:) If you want to apply the theorem quoted, then you just have to verify that $\Phi:M\to M\times N$, $x\mapsto (x,f(x))$ is an immersion, and it is a homeomorphism onto its image, which is the graph $\Gamma(f)$. I leave these two as exercises:)
