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My question is obviously based on the title. I want to show that there is no retraction of a $3$-sphere (denoted $S^3$) onto the torus $T^2$ (doughnut surface). Any ideas on how one should do this? Input would be highly appreciated.

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2 Answers 2

up vote 10 down vote accepted

If there were a retraction $r: S^3 \rightarrow T^2$ then the induced map $r_\ast: \pi_1(S^3) \rightarrow \pi_1(T^2)$ would be surjective. But $\pi_1(S^3)$ is trivial and $\pi_1(T^2)$ is not.

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I had an inkling that this had to do something with fundamental groups, thanks for the guidance. –  Libertron Jan 7 '13 at 1:33

Jacob Schlather's answer is probably the best. But, if you wanted a purely topological reason, you could use the fact that a $T^2$ in $S^3$ must bound a solid torus on at least one side. Then a retract of $S^3$ onto $T^2$ would induce a retract of $D^2\times S^1$ onto $T^2$, which is impossible (this last statement may require some justification).

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