Let $D_1,D_2$ be 2-dimensional discs, whose boundaries are $S_1$ and $S_2$. Paste $e^{2\pi it}$ in $S_1$ to $e^{2n\pi it}$ in $S_2$, and then we get a quotient space $X$. Prove that $X$ is simply connected.

If $n=1$, then $X=S_2$, so it's easy to imagine $X$ is path connected. But when $n\geqslant2$, the topology in X becomes complex, and I don't know how to construct a homotopy (because it needs to be continuous) between a loop and a constant loop.

Or should I use the fact that $S_2$ is simply connected?


1 Answer 1


For instance, since the subspace $D_1\subseteq X$ is contractible we have a homotopy equivalence $X\simeq X/D_1$. But $X/D_1 \equiv S^2$.

  • $\begingroup$ Thanks a lot. So I needn't to worry about the equivalence. $\endgroup$
    – Zeng
    Jun 5, 2018 at 20:15
  • $\begingroup$ Attaching is more interesting when the space you are attaching to is not contractible. If you attach a disk to $S^1$ along the map you described, the value of $n$ matters. $\endgroup$
    – user17892
    Jun 7, 2018 at 14:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .