(no) Continuous Injection from Torus $S^1 \times S^1$ into $S^2$ I'm trying to show there is no injection $f$ between the 2-torus $ S^1 \times S^1$ ( Henceforth " The Torus") and the 2-sphere $S^2$
My ideas so far:

*

*The Torus is compact, the 2-sphere is Hausdorff. Then $f: T^2 \rightarrow f(T^2) \subset S^2 $ is a homeomorphism, implying the Torus can be embedded in the 2-Sphere. not clear that's enough.(It's been a while since I saw this material. Hope I'm not too far off)

This question :No continuous injective function from 2-sphere to torus is similar but goes in the opposite direction


*As a map $f: T^2 \rightarrow S^2, f$ is an element of $\pi_2(T^2)$, which by Kunneth is trivial: ( Edit: As pointed out, this is not the case; rather the other way round; it's the group of homotopy classes of maps from S^n into X )

$\pi_2 (T^2)=\pi_2(S^1 \times S^1)= \pi_2(S^1)\times \pi_2(S^1)=0$ (*)
So that every map in $\pi_2(T^2)$ is homotopically trivial. But I can't tell if any such injective map violates this triviality.
Thanks for your suggestions.
Can this be done without Invariance of Domain? I vaguely remember it, but I never got a good feel for it.
*Up to some type of isomorphism.
 A: If you could embed the torus in the sphere, then --- since the sphere and the torus are not homeomorphic --- you could embed the torus in the plane. Now consider a parallel and a meridian in the torus, their images in the plane, and remember Jordan's theorem in curves.
A: Since $S^1\times S^1 \ncong S^2$ this embedding is not surjective and therefore $S^1\times S^1$ embeds in the plane $\mathbb{R}^2$. So we have a subspace of the plane with fundamental group $\mathbb{Z}^2$, which is finitely generated but not free. This is a contradiction. See https://math.stackexchange.com/a/2643521/1016538.
A: It's enough to prove that $T^2$ does not embed in $\mathbb R^2.$ But $T^2$ isn't even immersed in $\mathbb R^2.$ For, if so, then for an immersion $f,$  one could find a boundary point $v$ in the image of $f$, (by the compactness of $f(T^2)$). Now, take a $u\in f^{-1}(v)$ and consider $f_*:T_u(T^2)\to T_v(\mathbb R^2).$ The matrix of $f_*$ is the Jacobian $\mathscr J$ of $f$ at the point $u$ (relative to some charts $(U,\varphi), (V,\psi)$ about $u,v,$ respectively). If $\mathscr J$ were invertible, then the inverse function theorem would show that $f$ is a diffeomorphism between some neighborhoods $u\in U'$ and $v\in V'$ in $T^2$ and $\mathbb R^2$, respectively. But this is not possible, because $v$ is a boundary point of $f(T^2).$
A: As already mentioned in other answers it is enough to show that the torus does not embedd in the plane.
Supposing otherwise, assume there is such an embedding
$$f:\mathbb T^2 \to \mathbb R^2.
$$
Choose a point $p$ in the torus such that $f(p)$ lies in the boundary of $f(\mathbb T^2)$, and let $D$ be a disk in $\mathbb T^2$ centered at $p$.  Clearly $f$ restricts to an embedding of $D$ in the plane.
The restriction of $f$ to the boundary of $D$ is therefore a Jordan curve, whose complement in $\mathbb R^2$ is the union of a bounded, connected open set $U$ and an unbounded one, as in the Jordan curve theorem.
Since $f(p)$ is not in the Jordan curve, it must be in $U$ but this contradicts the fact that $p$ lies in the boundary of the range of $f$.
