Proof that a function $f:\mathbb{R}\times\mathbb{R}\to N$ restricts to a smooth function on $S^1$ I have to prove

Proof that a function $f:\mathbb{R}^2\to N$ (where $N$ is a smooth manifold) restricts to a
  smooth function on $S^1$

Here $S^1$ is defined as the subset of $\mathbb{R}^2$ satisfying $x^2+y^2=1$.
Now this it fairly trivial, but I want to be sure I understand what's involved in proving this. My idea is this:

Take $(a,b)\in S^1\subset \mathbb{R}^2$, and let $(V,\psi)$ be a chart around $(a,b)$.
If we take the atlas $\{(\mathbb{R}^2,\mathbb{1})\}$ for $\mathbb{R}^2$ and let $(U,\phi)$ be a chart around $f((a,b))$ then $\phi\circ f\circ \mathbb{1}^{-1}=\phi\circ f$ is smooth. Since $\psi$ is a diffeomorphism we thus conclude that $\phi\circ f\circ \psi^{-1}$ is smooth$\square$

I have the following $2$ questions:


*

*The author proves this claim by proving that the inclusion map $i:(a,b)\mapsto (a,b)$ is smooth on $S^1$, and hence $f|S^1=f\circ i$ is smooth. Is this necessary? I don't see the flaw in my proof, which seems to be much quicker.

*I find the claim that $f|S^1=f\circ i$ quite strange. Because the notation $f|S^1$ suggests we are thinking of $S^1$ as a real subset of $\mathbb{R}^2$, already included in the domain. However, if that is the case, then I don't see why we require to go through $i$ at all. If we think of $S^1$ 'on it's own'; not a literal subset of $\mathbb{R}^2$ and hence not included in the domain, then $f\circ i$ makes sense to me, but then $f|S^1$ doesn't make sense.


I hope you understand my confusion in part $2$. It might seem a bit overly nit-picky, but I feel that it is important to realise how we think about these things.
Addition to point $2$: A similar type of confusion happens when I (or the author of the book) use $\mathbb{R}^2$ with the smooth structure $\{(\mathbb{R}^2,\mathbb{1})\}$. I feel that if $f:\mathbb{R}^2\to N$, then that $\mathbb{R}^2$ is just the plain old euclidean plain, no structure. So why can we just say that all of a sudden $\text{dom}f$ is $\mathbb{R}^2$ with the smooth structure $\{(\mathbb{R}^2,\mathbb{1})\}$? To me these are $2$ different things. One is the 'normal, plain' euclidean plain, the other is a $2$ dimensional manifold looking a lot like the euclidean plane. 
 A: In your argument that $f|_{S^1}$ is smooth, you are actually implicitly using that $i : S^1 \hookrightarrow \Bbb R^2$ is smooth. Notice what we're doing when we write $\phi \circ f \circ \psi^{-1}$. The codomain of $\psi^{-1}$ is an open subset of $S^1$, while the domain of $f$ is $\Bbb R^2$. The fact that you've probably already seen is that if $f: M\to N$ and $g: N\to P$ are smooth maps between smooth manifolds, then $g \circ f: M \to P$ is smooth. Here we've assumed that the codomain of $f$ is the domain of $g$. Thus we can't immediately conclude that $f \circ \psi^{-1}$ is smooth from this fact since the hypotheses of the fact are not met.
On the other hand, simply slipping in the inclusion map $i$ solves this problem since, indeed, $f|_{S^1} = f \circ i$ and the codomain of $i$ is $\Bbb R^2$, which is the domain of $f$.
To see that $f|_{S^1} = f\circ i$, notice that since $i: S^1 \to \Bbb R^2$ and $f: \Bbb R^2 \to N$, when we compose we get $f \circ i: S^1 \to N$. Likewise $f|_{S^1} : S^1 \to N$, so the domains and codomains of these functions agree. Furthermore, it's very easy to see that for any point $(x, y) \in S^1$, $(f\circ i)(x, y) = f(i(x, y)) = f(x, y)$ and $f|_{S^1}(x, y) = f(x, y)$. Thus, these functions really are the same.
It's perfectly fine to think of $S^1$ without the ambient space $\Bbb R^2$, but you should keep the natural inclusion map in mind. It's a very useful function to think about when you're dealing with a space which has a natural ambient space, as spheres do.
