# Meaning of 'each of which is contained in a single $A_{\alpha}$,' in Hatcher's Algebraic Topology

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!

• I reckon it's "there exists an $A_\alpha$" not "there is a unique \$A_\alpha". – Lord Shark the Unknown Apr 21 '17 at 7:01
• 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. – Pierre Apr 21 '17 at 7:34

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.