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Lemma 1.15. of Hatcher's Algebraic Topology says: if a space $X$ is the union of a collection of path-connected open sets $A_{\alpha}$ each containing the base point $x_0\in X$ and if each intersection $A_{\alpha}\cap A_{\beta}$ is path-connected, then every loop in $X$ at $x_0$ is homotopic to a product of loops each of which is contained in a single $A_{\alpha}$.

I am wondering what exactly is the meaning of 'each of which is contained in a single $A_{\alpha}$.' Does it mean that for each loop in the product, there exists a $A_{\alpha}$ that fully contains it? Or does it mean that for each loop in the product, there is a unique $A_{\alpha}$ that fully contains it? The distinction being that in the first case, a loop in the product might be a subset of $A_{\alpha}\cap A_{\beta}$ whereas this is excluded in the second case.

I am struggling to follow the associated proof and I think that clearing up what we are trying to show would help me out. Furthermore, I have been told that it is important to understand this Lemma because it leads to van Kampen's Theorem. Thanks!

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    $\begingroup$ I reckon it's "there exists an $A_\alpha$" not "there is a unique $A_\alpha". $\endgroup$ – Lord Shark the Unknown Apr 21 '17 at 7:01
  • $\begingroup$ I've thought about it and I think I agree. I think the proof makes sense if that's what we're trying to show. But I'm not super confident so if someone would care to confirm with some certainty, that'd be great. $\endgroup$ – Pierre Apr 21 '17 at 7:34
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I had a doubt in understanding the proof of the above-mentioned lemma, which was also due to the term, "single $A_{\alpha}$", being used. In particular, the part where Hatcher mentions that $V_s$ can be taken to be an interval whose closure is mapped to a single $A_{\alpha}$.

My attempt to solve it is was follows: Let $s \in I$ such that $f(s) \in A_\alpha \cap A_\beta$ for some $\alpha, \beta$. Now, consider some open neighbourhood $V_s$ in $I$ which is mapped to some $A_\alpha$. For the sake of simplicity, let $V_s=(s_i,s_{i+1})$. Now, consider $(s_i,s)$ where $f(s) \in V_\alpha \cap V_\beta$. As the closure of ${(s_i,s)}$ is $[s_i,s]$, it is mapped to a single $A_{\alpha}$. Note $f([s_i,s]) \subset A_\alpha \cap A_\beta$ but $f([s_i,s]) \cap A^c_\alpha = \phi$.

Hence $V_s$ can be taken to be an interval whose closure is mapped to a single $A_{\alpha}$. I hope it clears your doubt.

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