# The Torus is not homeomorphic to the unit circle

Let $$T^2=S^1 \times S^1$$ be a torus in $$\mathbb{R}^4$$, where $$S^1:=\{x \in \mathbb{R}^2 \mid \|x\|_2 = 1\}$$ is the unit circle. Does there exist a homeomorphic map $$f: T^2 \to S^1$$ ?

Intuitively this makes sense (somewhat), since $$\mathbb{R}^4 \cong \mathbb{C}^2$$ can be seen by considering the isomorphism $$\mathbb{R}^4 \to \mathbb{C}^2, \; (a,b,c,d) \mapsto (a+\mathrm{i}b,c+\mathrm{i}d)$$ but I wonder whether it is possible to explicitely write down a homeomorphism $$T^2 \to S^1$$. I know that a homeomorphism is a continuous bijection with a continuous inverse map.

If not, I'd be interested in knowing how to start tackling this problem, because right now, I'm clueless as to where to start.

• Torus and circle are not homeomorphic. Fondamental group of the circle is $\mathbb Z$ whereas fundamental group of the torus is $\mathbb Z^2$.
– Surb
Mar 2 '21 at 11:29
• The claim is completely false: if you remove any two points from $T^2$ you obtain a connected space, whereas whatever two distinct points of $S^1$ you remove, you obtain a disconnected space (and all its connected subsets with at least two points have this property, therefore $T^2$ can't even be embedded in $S^1$).
– user239203
Mar 2 '21 at 11:31
• The torus is 2-dimensional; the circle is 1-dimensional. Dimension is a topological invariant. Mar 2 '21 at 15:01

It is well-known that $$\pi_1(S^1)\cong\Bbb Z$$. From this it follows by functoriality that $$\pi_1(S^1\times S^1)\cong\pi_1(S^1)\times\pi_1(S^1)\cong\Bbb Z\times\Bbb Z\cong\Bbb Z^2$$.