Confused with dimensions and embeddings I'm new to topology and apologize in advance for this, perhaps, very simple (or philosophical) question.
I've always thought of a torus as a doughnut-shaped surface in $\mathbb{R}^3$. However, after I began studying topology I've found out that torus is $S^1 \times S^1$ and it is naturally defined in $\mathbb{R}^4$. But at the same time, as I understood, popular 3d representation of a torus is an embedding in $\mathbb{R}^3$, so, by definition of embedding, natural 4d torus is homeomorphic to easily visualized 3d torus.
When we take the quotient of a square (by identifying sides) to construct a torus, aren't we tricking ourselves visualizing this in $\mathbb{R}^3$, since we just get some "slice" of a real 4d torus. I may have answered my own question here by stating that embedding is a homeomorphism, but I still want to understand what are the connections between dimension, embedding and homeomorphism.
Torus is 2-dimensional, since 2 points are enough to define it (one point for each $S^1$), but each circle is naturally presented in $\mathbb{R}^2$, thus we need $\mathbb{R}^4$.
Are we losing "information" when we "project" the torus from $\mathbb{R}^4$ to $\mathbb{R}^3$? Is it only visual loss or also topological?
I can imagine taking 3-ball in $\mathbb{R^3}$ and "shrinking" it to a 2-ball (disk) in $\mathbb{R}^2$ by $z \to 0$. During this transition from $\mathbb{R}^3$ to $\mathbb{R}^2$ we obviously lost both visual and topological information (n-ball is homeomorphic to m-ball $\iff$ n=m).
Does homeomorphism preserve "inner" dimension, but doesn't "care" about outer (extrinsic) space?
 A: I don't really view the 'natural' torus as $S^1 \times S^1$ sitting in $\mathbb{R}^4$. There are multiple equivalent (read: homeomorphic) ways of seeing the torus, one of which is the familiar 'donut' picture. Two other ones would be as $S^1 \times S^1$ sitting in $\mathbb{R}^4$, or as a quotient of the square, as you indicated.
The bottom line is that to a mathematician, the torus is an object in its own right. Whether there exists an ambient Euclidean space into which you can embed it is in some sense irrelevant. It's just a set of points together with a collection of 'open subsets' which define its shape.
To come to your questions: given a topological space (for example, the space $X$ which is quotient of the square by identifying opposite sides carrying the quotient topology), we can try to visualize it by embedding it into a Euclidean space. An embedding of the topological space $X$ into Euclidean space $\mathbb{R}^n$ is just a map $\phi: X \rightarrow \mathbb{R}^n$ such that $\phi: X \rightarrow \phi(X)$ is a homeomorphism.
So, it turns out that $X$ can be embedded into $\mathbb{R}^3$, but also in $\mathbb{R}^4$. Think of these as 'realizations' of $X$ in some larger ambient space. Both these realizations are homeomorphic to $X$ (duh, by definition of what an embedding is), so they are also homeomorphic to each other. Thus, no information is lost.
It is not correct to think of the 'donut' picture of the torus as a projected version of the realization in $\mathbb{R}^4$. There is no projection going on (like when you project a vertical cylinder in 3D to a circle slice in the horizontal plane). The donut is not a 3D slice of the 4D shape, it's the same shape.
You are correct to say that the dimension of the torus is $2$. This dimension is also independent of the ambient space. Homeomorphism therefore preserves this dimension, and doesn't care about extrinsic dimension. There is a bit of a caveat here: it's pretty hard to define what 'dimension' means for a topological space, so proving the claim that the torus has dimension 2 is hard.
