# Question regarding a remark in Hatcher's Algebraic Topology

I am having difficulty understanding the following remark made by Hatcher on page 50 of Algebraic Topology:

My understanding of this paragraph is as follows: For a basepoint $s_0\in S^1$ the attaching map $\varphi_{\alpha}$ determines a loop based at $\varphi_{\alpha}(s_0)$ in $Y$, namely the boundary of the image of $e^2 _{\alpha}$. For some path $\gamma_{\alpha}$ from $x_0$ to $\varphi_{\alpha}(s_0)$ one can use the change of base isomorphism, one sees that the loops $\varphi_{\alpha}(s_0)$ can be associated with loops all based at the same point $x_0 \in X$. This is as far as I can follow. I don't understand why although the loop $\gamma_{\alpha}\varphi_{\alpha}(s_0)\overline{\gamma_{\alpha}}$ will be nullhomotopic after attaching $e^2_{\alpha}$, even though it may not be before.

Why is this? Also, are there examples which can elucidate why this is true?

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The simplest example would be taking a generator of $\pi_1(S^1,x_0)$, pick any $x_0$, and attach a single 2-cell to $S^1$ in the normal way. Then the (previously non-zero) loop becomes null-homotopic, since this new space is just $S^2$(EDIT: $D^2$, as pointed out in the comments).

In general, the loop $\varphi_\alpha$ will become null-homotopic, since we can just 'shrink' it to a constant map in the 2-cell we just attached(via, e.g. straight line map in the disk), $e^2_\alpha$. So imagine in your head you have your loop $\gamma_\alpha \varphi_\alpha \overline{\gamma}_\alpha$, now mentally 'shrink' $\varphi_\alpha$ so we're now homotopic to $\gamma_\alpha \overline{\gamma}_\alpha$, but that is obviously nullhomotopic.

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Attaching a $2$-cell to $S^1$ in the normal way seems to give a space homeomorphic to $D^2$, not $S^2$, though this would be homeomorphic to a hemisphere. –  Holdsworth88 Sep 5 '12 at 14:52
Thank you, I mis-wrote it –  John Stalfos Sep 5 '12 at 15:12
Let $X$ be an annulus with boundary and let $\phi_\alpha$ be the map that maps $S^1$ homeomorphically to the inner boundary of the annulus. Pick any point $x_0$ in the interior of the annulus and paths from $x_0$ to $\phi_\alpha(s_0)$. Then $\gamma_\alpha\phi_\alpha \bar{\gamma}_\alpha$ is certainly not nullhomotpic as it is in the equivalence class of the generator of the fundamental group of the annulus ($\mathbb{Z}$).
After the attaching is performed, the space $Y$ is a disk, as the map $\phi_\alpha$ attaches the two-cell and fills in the center of the annulus. Therefore, after attaching the disk, the loop becomes nullhomotopic.