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Does there exists a continuous surjective map from $S^1 \times S^1$ to $S^2$ ?

Silambarasan C
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    Welcome to Math SE. Take a look at some questions and answers and you will find that ones that get a better reception and responses are ones where the OP explains not only the context of the question, but shows their thought process and/or attempts and where they got stuck. Context is important, e.g. since this is only tagged as general topology, useful connections to algebraic topology may not be provided. – AlgTop1854 Nov 06 '23 at 16:00
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    Can we please give a new contributor a chance to absorb feedback and revise before TWO downvotes? – AlgTop1854 Nov 06 '23 at 16:03
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    such as projection onto a sphere that is completely inside the torus? – Hagen von Eitzen Nov 06 '23 at 16:11
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    Welcome to (posting at) MSE! <> This is stronger than what you ask, but the diagram may help you visualize one way to get such a mapping. – Andrew D. Hwang Nov 06 '23 at 16:47

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Consider the map $f:S^1\times S^1 \to S^2$ by $(e^{i \theta},e^{i \phi})$ $\to$ $(\cos\theta \sin \phi,\sin \theta \cos \phi, \cos\phi)$. The above map is continuous because it is coordinate wise continuous and surjective because it is nothing but spherical parametrization of unit sphere.

diracdeltafunk
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simbu naveen
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Another way to see this: view $S^1 \times S^1$ as a quotient of $[0,1] \times [0,1]$ (the standard picture with opposite sides of the square identified). There is a further quotient given by collapsing the entire boundary of $[0,1] \times [0,1]$ to a point. This further quotient is $S^2$.

diracdeltafunk
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Similar to the prior answer and in the spirit of how this might be thought of by algebraic topologists:

First take the torus and pinch one factor $S^1$ to a point (this turns the bagel into a hot dog where the tip of each end is joined), which has the homotopy type of $S^1\vee S^2$. Then collapse $S^1$ to a point, resulting in $S^2$. The sequence looks like this where each arrow is a quotient map:

$S^1\times S^1\rightarrow S^1\times S^1/S^1\times \{*\}\simeq S^1 \vee S^2\rightarrow S^2$

AlgTop1854
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