I haven't learned the concept of "transverse" yet, and it just appears in the question.
Consider two maps $$\begin{align} f_{0}:U_{0}\to \mathbb{R}^n \\ f_{1}:U_{1}\to \mathbb{R}^n \end{align}$$ and with $U_{0}\subset \mathbb{R}^{k_{0}}; \ U_{1}\subset \mathbb{R}^{k_{1}}$ are $2$ open sets. We say $f_{0},f_{1}$ are transverse if for all $x\in U_{0}, \ y\in U_{1}$ , with $$\begin{align} f_{0}(x)=f_{1}(y)=p\in \mathbb{R}^n \end{align}$$ we have that any vector $v\in \mathscr{T}_{p}(\mathbb{R}^n)$ can be expressed as a sum $$\begin{align} (p;v)=(p;Df_{0}(x)v_{0})+(p;Df_{1}(y)v_{1}) \end{align}$$ for $v_{0}\in \mathscr{T}_{x}(U_{0})$ and $v_{1}\in \mathscr{T}_{y}(U_{1})$ .
I'm asked to solve the following problems.
1 Let $$\begin{align} \alpha :t \to (\cos t, \sin t) \end{align}$$ and for each $r \in \mathbb{R}$ , $$\begin{align} f_{r}:t\to (t,r) \end{align}$$ I'm asked to prove $\alpha$ and $f_{r}$ are transverse if and only if $r\ne \pm 1$
2 $f_{0}:U_{0}\to \mathbb{R}^n \ \ \ f_{1}:U_{1}\to \mathbb{R}^n$ , and consider the set $$\begin{align} M=\{ (x,y)\in U_{0} \times U_{1}:f_{0}(x)=f_{1}(y) \} \end{align}$$ Prove $M$ is a manifold without boundary.
For the first one, I have no idea to prove the statement. For one direction, if they are transverse, this means that $$\begin{align} (p;v)=(p; D\alpha(t)u_{1})+(p;Df_{r}(t')u_{2})= (p;\begin{pmatrix} -\sin t \\ \cos t \end{pmatrix}u_{1})+(p;\begin{pmatrix} 1 \\ 0 \end{pmatrix}u_{2}) \end{align}$$ where $u_{1},u_{2}$ are 2 numberse from $\mathbb{R}$. I can't see how to get $r \ne \pm 1$ from here. As for the other direction, I have no idea.
For the second one, I consider the map $g(x,y)=f_{0}(x)-f_{1}(y)$, and $M=\{ (x,y):g(x,y)=0 \}$ . According to the theorem, I know, as long as I can show $M$ is not empty and $\text{rank }Dg(x,y)=n$, I can prove $M$ is a $n$ dimensional manifold without boundary. However, I don't know how to show the rank as $n$. I do know that $$\begin{align} Dg(x,y)=(Df_{0}(x) \ \ -Df_{1}(y)) \end{align}$$ which is a $n \times k$ matrix. Here $k=k_{0}+k_{1}$. I think in order to show the rank is $n$, I should use the property that $f_{0},f_{1}$ are transverse.
Any help on this? Thanks!