Sphere is a topological manifold. If I want to show that the spere $\mathbb{S}^1$ is a topological manifold. Graphically it's clear that we have a chart $x:\mathbb{R}\rightarrow \mathbb{S}^1$ since we can "cut" $\mathbb{S}^1$ and and form it to a line right? But somehow I can't imagine how to show this mathematically.
To have the context, I want to show the following statement

The n-dimensional torus $\mathbb{T}^n \subset \mathbb{C}^n$ with it's subspace topology is a topological manifold of dimension n.

I wrote $\mathbb{T}^n=\mathbb{S}^1\times ...\times \mathbb{S}^1$ as an n-dimensional product. Then i thought if I could show that $\mathbb{S}^1$ is a topological manifold, also the product is one.
Another question is, why is the subspace topology so important in this case, don't I need to use the product topology since I rewrote the torus as a product?
thank you!
 A: If you define $T^n$ as the product of circles, then it has the product topology and no reference to the subspace topology is necessary. Note, however, that both topologies agree (which is a little lemma: If you have subspaces $A_i \subset X_i$, then $\prod_i A_i$ with the product topology is a subspace of $\prod_i X_i$). Here you have $S^1 \subset \mathbb C$.
If you know that $S^1$ is a $1$-manifold, then of course you also know  that $T^n$ is an $n$-manifold. See your last question.
To show that $S^1$ is a manifold, you can use stereographic projection (see here) or the following four charts (where $J  = (-1,1)$):
$$a_1 :  \{ z \in \mathbb C \mid \Re (z) > 0 \} \to J, a_1(z) = \Im(z)$$
$$a_2 : \{ z \in \mathbb C \mid \Re (z) < 0 \} \to J, a_2(z) = \Im(z)$$
$$a_3 : \{ z \in \mathbb C \mid \Im (z) > 0 \} \to J, a_3(z) = \Re(z)$$
$$a_4 : \{ z \in \mathbb C \mid \Im (z) < 0 \} \to J, a_4(z) = \Re(z)$$
You can translate this to your chart concept by considering $a^{-1}_i$ and composing with a homeomorphism $h : \mathbb R \to J$.
