# Prove a quotient space which consistis of 2 discs is simply connected.

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

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

• Thanks a lot. So I needn't to worry about the equivalence.
– Zeng
Jun 5, 2018 at 20:15
• 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.
– user17892
Jun 7, 2018 at 14:19