Topologies on the Torus

Question

In the text "Lecture notes on Elemental Topology and Geometry" by Singer & Thorpe the following is stated:

"The the unit circle $S^1$ in $\mathbb{R^2}$ and can be made into a topological space with the subspace topology (here $\mathbb{R^2}$ has the usual topology).

Also $S^1\times S^1$ is a torus.

The student should verify that the product topology on the torus is the same as the subspace topology of the torus considered as a subset of $\mathbb{R^3}$."

Is this correct? shouldn't it be $\mathbb{R^4}$?

I know that the product topology is generated by sets of the form $U\times V$ where $U,V$ are open in $S^1$.

Can you give me any hints?

Thanks in advance

Answer 1

If $S^1$ is considered to be a subspace of $\mathbb{R}^2$, then $S^1\times S^1$ is a subspace of $\mathbb{R}^4.$

The authors are also referring to the conventional "surface of a donut" torus, which is a subspace of $\mathbb{R}^3.$

They're asking you to show that topologies of these two objects are the same.

Answer 2

The product (subspace) open sets $U_1 \times U_2$ ( arcs of the circle) are 2-dimensional and can be embedded in $\mathbb R^3 $ or higher. You could , of course consider the embedding in $ \mathbb R^4$-or-higher, but you can also consider embeddings in $\mathbb R^3 $. So you need to show that basic open sets $U_1 \times U_2 $ , or cylinders within the torus are open in the subspace topology, i.e., each cylinder can be obtained by intersecting a set open in $\mathbb R^3 $ with the torus. Consider , e.g., an open cylinder living in $\mathbb R^3$ . This shows the Torus cylinders are open in the Subspace topology . Now show the opposite direction and you are done.