I'm trying to solve the following problem. Some help and feedback would be helpful.
Suppose $f: S^1 \to \mathbb{R}^2$ and $g: S^1 \to \mathbb{R}^2$ are smooth embedding. Let $$M = \{(a, b, \overrightarrow{v})\in S^1 \times S^1 \times \mathbb{R}^2: f(a) - g(b) = \overrightarrow{v}\}$$
Show that $M$ is a compact submanifold $S^1 \times S^1 \times \mathbb{R}^2$. Let $\pi: M \to \mathbb{R}^2$ be the projection $\pi(a, b, \overrightarrow{v} )= \overrightarrow{v}$. Apply Sard's Theorem to $\pi$ and deduce that for almost every $\overrightarrow{v} \in \mathbb{R}^2$, $f(S^1)$ is transverse to $g(S^1) + \overrightarrow{v}$
My incomplete solution is as follows:
For the first part I want to use the regular value theorem, meaning $M = \pi^{-1}$ of something but I'm not sure what that something is. For compactness, I want to use closed and boundedness but since $v$ can be any vector in $\mathbb{R}$ I don't think it's bounded.
For the second part, let the set of points $X$ such that the differential of $\pi$ has deficient rank has an image of measure $0$. Then by Sard's theorem, the measure of $\pi(X)$ is $0$. The differential of $\pi$ is $d\pi = \begin{bmatrix}df & -dg \end{bmatrix}$ Unsure about this computation so that the set of points $X$ such that $d\pi$ has deficient rank are the set of points such that $df$ and $dg$ have the deficient rank as well. If $f(S^1)$ is not transverse to $g(S^1) + \overrightarrow{v}$, then $d\pi$ at these points has deficient rank. not sure why this last part should be true if it is even true
The obvious mapping to use is
$$ F : \mathbb S^1 \times \mathbb S^1 \times \mathbb R^2\to \mathbb R^2, F (a, b, \vec v) = f(a) - g(b) -\vec v$$ and $M = F^{-1} (\vec 0)$. To show that $M$ is an embedded submanifolds, it suffices to show that $dF_x$ is surjective for all $x\in M$. But indeed $dF_x$ is surjective for all $x=(a, b, \vec v)$: since the last action of $F$ on $\vec v$ is affine,
$$\tag{1} dF_x (t_1, t_2, \vec w) = t_1f'(a) - t_2 g'(b) + \vec w.$$ In particular, $$ dF_x (0,0,\vec w) = \vec w$$ for all $\vec w\in \mathbb R^2$ and hence $dF_x$ is surjective.
To show that $M$ is compact, note that closedness is obvious since $\pi$ is continuous. On the other hand, if $(a, b, \vec v)\in M$, then
$$ |\vec v| \le |f(a)| + |g(b)| \le M_1+M_2,$$ where $M_1, M_2$ depend only on $f, g$ respectively. Thus $M$ is bounded.
Lastly, consider the smooth map $\pi : M \to \mathbb R^2$, $\pi(a, b,\vec v) = \vec v$. The Sard theorem tells us that there is a measure zero set $S \subset \mathbb R^2$ so that for all $\vec v\notin S$ and for all $x=(a, b, \vec v) \in M$,
$$ d\pi_x : T_xM \to \mathbb R^2$$
is surjective. Note that $T_xM = \ker dF_x$. From (1),
$$ \tag{2}T_xM = \{ (t_1, t_2, \vec w) \in \mathbb R^4: t_1f'(a) - t_2 g'(b) + \vec w=0\}, $$
Hence $d\pi_x : T_xM \to \mathbb R^2$ is given by
$$d\pi_x (t_1, t_2, \vec w) = \vec w = -t_1 f'(a) + t_2 g'(b).$$
by (2). That is, $d\pi_x$ is surjective if and only if $f'(a), g'(b)$ are linearly independent. This is the same as that $f(\mathbb S^1), g(\mathbb S^1)+\vec v$ intersects transversally at $f(a)= g(b)+\vec v$.