Interior of a compact 3-manifold I have an orientable 3-manifold $X$, such that
 $$X=\lbrace(x,y,z)\mid x\neq y \neq z \neq x \rbrace\subseteq S^1\times S^1 \times S^1 $$
How to find a compact 3-manifold $M$ such that $X= interior(M)$
any help please???
 A: I believe your desired manifold with border $M$ are two disjoint, solid, 2-tori.
First, let's have a look at the coordinates: We can parametrize $\mathbb{S}^1$ as $ℝ/ℤ$. Next we choose new coordinates $(a,b,c)$ on the 3-torus $\mathbb{S}^1×\mathbb{S}^1×\mathbb{S}^1$.
$$(a,b,c) = φ(x,y,b) = (x, y-x, z-y)$$
$$(x,y,z) = φ^{-1}(a,b,c) = (a, a+b, a+b+c)$$
We see that $φ$ is a homeomorphism from $ℝ^3$ to $ℝ^3$ and both $φ$ and $φ^{-1}$ map the subset $ℤ^3$ into $ℤ^3$. It follows that this defines an homeomorphism from $T^3 = ℝ^3/ℤ^3$ to itself.
What happens to our equations?


*

*$x \neq y ⇔ a \neq a + b ⇔ b \neq 0$

*$y \neq z ⇔ a+b \neq a + b+c ⇔ c \neq 0$

*$x \neq z ⇔ a \neq a + b+c ⇔ b+c \neq 0$


$$X \simeq \{(a,b,c)|b\neq0, c\neq0, b+c\neq0\} \subset (\mathbb S^1)^{×3}$$
$$\{(a,b,c)|b+c\neq1\} \subset \mathbb S^1 × (0,1)^2$$
$$\mathbb S^1 × \{(b,c)\in (0,1)^2|b+c\neq1\}$$
Which is a lot easier to handle. It is OK to change the condition to $b+c \neq 1$, as that is the same modulo $ℤ$ and one happens to be the representant of $0$ which is actually in range. 
Now $\{(b,c)\in (0,1)^2|b+c\neq1\}$ is just a set of two triangles which almost  touch each other. That almost touching isn't registered by the manifold
structure on $X$. Separating the triangles and taking the closure yields
$$X = \text{interior}(\mathbb S^1 × (\{(b,c) \in [0,1]^2|c \leq b\}\sqcup\{(b,c) \in [0,1]^2|c \geq b\}))$$
Side note: The analogous problem on the $2$-torus gives one strip $\mathbb S^1 × [0,1]$.
A: This is a variation on Rolf Sievers's answer, written to translate his picture into something more familiar to me.  So, if you like my answer, you should also upvote Rolf's answer!
The first thing to note, is that your space $X$ is nothing but the configuration space of $3$ points on a circle, $F(S^1, 3)$.  That is, one way to picture your space $X$ is as three points on $S^1$ which are free to move, but not touch each other.  This, for example, allows one to easily see that $X$ is disconnected.  For if we label the three points $x,y,z$, then by suitable rotations, we may assume $x = (0,1)$.  If we head clockwise from $x$, we either encounter $y$ first or $z$ first - these two possiblilities gives rise to the two connected components of $X$.
Now, one has the following result, which I first found in Fred Cohen's notes.

Suppose $G$ is a topological group and $F(G,k)$ denotes the configuration space of $k$ points in $G$.  Then there is a natural homeomorphism $F(G,k) \cong G \times F(G\setminus\{e\}, k-1)$

The proof is straight forward:  given $(g_1,..., g_k)\in F(G,k)$, map it to $(g_1, g_1^{-1} g_2, ..., g_1^{-1} g_k)$.  Given a point $(g, g_1,..., g_{k-1})\in F(G\setminus\{e\}), k-1)$, map it to $(g, gg_1, gg_2,..., gg_{k-1})$.  One easily sees these maps are inverses to each other.
As an added bonus, if $G$ is a Lie group, then the map we just wrote down is not just a homeomorphism, but also a diffeomorphism.
Now, we apply that to your example.  $S^1$ is Lie group, so we see that $F(S^1, 3) = S^1 \times F(S^1\setminus \{e\}, 2)$.
But $S^1\setminus \{e\}$ is homeomorphic to $\mathbb{R}$ (and also to $(0,1)$), which, in particular, is a topological group!  So, $F(S^1\setminus\{e\}, 2) \cong \mathbb{R} \times F(\mathbb{R}\setminus \{e\}, 1)$.
Finally, we note that for any space $Y$, $F(Y,1) = Y$.  So putting this all together, these space you care about is $$X = F(S^1, 3)  \cong S^1\times (0,1) \times [(0,1)\coprod (0,1)]$$ which is homeomorphic to two disjoint copies of $S^1\times (0,1)^2$.
As $(0,1)^2$ is homeomorphic to an open disc of radius $1$, each copy of $S^1\times (0,1)^2$ is the interior of $S^1\times D^2$, just as Rolf claimed.
