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I have to (show or) disprove that there is no continuous surjection from $D^2$ (unit closed disk of $\mathbb{R}^2$) to $S^1$ (unit circle)?

I would like to say that if $f: X \to Y$ is surjective, then $f_*: \pi_1(X) \to \pi_1(Y)$ is surjective too, which would imply the result in our case since $\pi_1(D^2)$ is trivial, but Im not sure if this is correct?

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  • $\begingroup$ No, that's not correct. In fact, there are surjections from the disk to the circle. $\endgroup$
    – Steve D
    Jun 13, 2018 at 4:30

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You can surject $D^2$ to a closed interval, then surject the closed interval to $S^1$. An easy formula: $(x,y)\mapsto (\cos10x,\sin10x)$.

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  • $\begingroup$ thanks. I was looking in the wrong direction! $\endgroup$
    – Thomas
    Jun 13, 2018 at 4:36

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