I'm going to assume that you know all about quotient spaces (if you don't then you'll have to learn that because otherwise no answer to your question will make sense).
Another thing you need is the collar neighborhood theorem from differential topology, applied to the boundaries of $X_1$ and $X_2$. That theorem says that there exist neighborhoods $N_1$ of $\partial X_1$ and $N_2$ of $\partial X_2$, and diffeomorphisms $f_i : N_i \to \partial X_i \times [0,1)$ (for each $i=1,2$) such that $f(x) = (x,0)$ for each $x \in \partial X_i$ .
Now choose a diffeomorphism $g : \partial X_1 \to \partial X_2$. And then we have the quotient topological space $Y$ together with the quotient map $q : X_1 \cup X_2 \to Y$ obtained by identifying each $x \in \partial X_1$ with $g(x) \in \partial X_2$, so $q(x)=q(g(x))$.
Let me alter your notation slightly: I'll choose $x_1 = \partial D_1 = \partial X_1$ and $x_2 = g(x_1) \in \partial D_2 = \partial X_2$, which corresponds to the point $x = [x_1] = [x_2] \in Y$ (here I use the notation $[\cdot]$ to denote the corresponding point in the quotient space; I'm unsure whether this is what you intend in your question when you put the $\tilde{}$ symbol over something).
So now I have to describe a manifold chart in $Y$ for the point $x$. To do this, I'll choose a manifold chart in $\partial X_1$ around $x_1$, i.e. an open subset $U_1 \subset \partial X_1$ containing $x_1$ and a diffeomorphism $\phi_1 : U_1 \to B$, where $B$ is the unit open ball in Euclidean space. From this I get a manifold chart in $\partial X_2$ around $x_2$, namely $U_2 = g(U_1)$ and $\phi_2 = \phi_1 \circ g^{-1} : X_2 \to B$.
In $Y$ define the open chart around $x$ as follows:
$$U = q\bigl(f_1^{-1}(U_1 \times [0,1)) \cup f_2^{-1}(U_2 \times [0,1))\bigr)
$$
The map $\psi : U \to B \times (-1,+1)$ will be given by the following formula. Each $y \in U$ has one of two forms, and we give the formula for each form:
- If $y = qf_1^{-1}(z,t)$ for $(z,t) \in B \times [0,1)$ then $\psi(y) = (z,-t) \in B \times (-1,0]$
- If $y = q f_2^{-1}(z,t)$ for $(z,t) \in B \times [0,1)$ then $\psi(y) = (z,t)$
This is well-defined if $y$ has both forms (which only happens when $t=0$).
And then, by tracing through all the definitions and using all the theorems you can possibly find from quotient spaces, it follows that $\psi$ is a homeomorphism from $U$ to $B \times (-1,+1)$.
No. Any compact $C \subset \mathbb R^k$ has the property that each $c \in C$ has an open neigborhood in $C$ that can be embedded in $\mathbb R^k$. Actually we may take any open neigborhood $U$ of $c$ in $C$ since $U \subset \mathbb R^k$.
I think you mean that each $c \in C$ has an open neighborhood which can embedded as an open subset of $\mathbb R^k$. Then in fact $C$ is a $k$-manifold. But of course you cannot expect that that all open neigborhoods are homeomorphic to $\mathbb R^k$. There are such "nice" neighborhoods, but also other neigborhoods (just remove a compact subset from a nice neighborhood).
Update on request:
Let $C$ be a compact space such that for each $c \in C$ there exists an open neighborhood $U_c \subset C$ and an embedding $\phi_c : U_c \to \mathbb R^{k(c)}$ for some $k(c)$. We shall construct an embedding $h : C \to \mathbb R^N$ for a sufficiently large $N$. Since $C$ is compact Hausdorff, hence normal, we find open $V_c \subset C$ such that $c \in V_c \subset \overline {V_c} \subset U_c$ and a continuous Urysohn-function $u_c : C \to [0,1]$ such that $u_c(c) = 1$ and $u_c(x) = 0$ for $x \in C \setminus V_c$. Define
$$g_c : C \to \mathbb R^{k(c)} \times \mathbb R = \mathbb R^{k(c)+1} , g_c(x) = \begin{cases} (u_c(x) h_c(x), u_c(x) )& x \in \overline {V_c} \\ (0, u_c(x)) = (0,0) & x \in C \setminus V_c \end{cases}$$
Both parts of the definition yield continuous functions on the closed subsets $\overline {V_c}$ and $C \setminus V_c$ of $C$, respectively. On $\overline {V_c} \cap (C \setminus V_c) = \operatorname{bd} V_c$ both parts agree, therefore $g_c$ is continuous.
Let $W_c = u_c^{-1}((0,1])$. This is an open subset of $C$ such that $c \in W_c \subset V_c$.
Claim: If $g_c(x) = g_c(y)$ and $x \in W_c$, then $x = y$.
Proof. $g_c(x) = g_c(y)$ implies $u_x(x) = u_x(y)$. For $x \in W_c$ we have $u_c(x) > 0$, thus also $u_c(y) > 0$ which shows $y \in W_c$. Hence $x,y \in W_c \subset V_c$ and thus $u_c(x) h_c(x) = u_c(y) h_c(y)$. Since $u_c(x) = u_c(y) > 0$, we conclude that $h_c(x) = h_c(y)$ and therefore $x = y$.
Since $C$ is compact, finitely many $W_{c_i}$ cover $C$. Define
$$h : C \to \prod_{i=1}^n \mathbb R^{k(c_i)+1} = \mathbb R^N, h(x) = (g_{c_1}(x), \ldots, g_{c_n}(x)) .$$
We shall check that $h$ is injective; thus it is an embedding because $C$ is compact. So let $x, y \in C$ such that $h(x) = h(y)$. We have $x \in W_{c_i}$ for some $i$, thus $g_{c_i}(x) = g_{c_i}(y)$ with $x \in W_{c_i}$. Our above claim shows that $x = y$.
Best Answer
Perhaps the equivalence relation you want is "isotopic", which briefly means "homotopic through homeomorphisms". More precisely, you want to know whether there is an isotopy from $C_0$ to $C_2$, which by definition is a continuous function $H : C_0 \times [0,1] \to \mathbb{R}^3$ such that $H(x,0)=x$, $H | C_0 \times \{t\}$ is a homeomorphism onto its image for each $t \in [0,1]$, and $H(C_0)=C_2$.
The boundary of $C_m$ when $m$ is even is a 2-component link which I'll denote $L_m$. As it turns out, given unequal even integers $m,n$ the links $L_m,L_n$ are not isotopic, and therefore the twisted cylinders $C_m,C_n$ are not isotopic. The reason is that for any 2-component link there is an isotopy invariant called the linking number. One can compute for each even number $m$ that the linking number of $L_m$ equals $m/2$ (see the picture on that web page which shows $C_8$).