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.

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    $\begingroup$ Torus and circle are not homeomorphic. Fondamental group of the circle is $\mathbb Z$ whereas fundamental group of the torus is $\mathbb Z^2$. $\endgroup$
    – Surb
    Mar 2 '21 at 11:29
  • 6
    $\begingroup$ 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 $S^1$). $\endgroup$
    – user239203
    Mar 2 '21 at 11:31
  • $\begingroup$ The torus is 2-dimensional; the circle is 1-dimensional. Dimension is a topological invariant. $\endgroup$ 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 $\pi_1(S^1\times S^1)\cong\pi_1(S^1)\times\pi_1(S^1)\cong\Bbb Z\times\Bbb Z\cong\Bbb Z^2$.

Since they have different fundamental groups, they can't be homeomorphic.


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