# Where do we use path-connectedness in the proof of van Kampen's theorem?

I have a question about the proof of Van Kampen's theorem in Hatcher's book, which is theorem 1.20 on p43) (see here: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf)

Where exactly in the proof of this theorem does one use that triple intersections remain path connected? It seems that Hatcher critically uses this to show that $$\gamma_0$$ and $$f_1* \dots* f_k$$ have equivalent factorisations (and similarly for $$\gamma_{mn}$$ and $$f_1' * \dots * f_l'$$).

Do we need this earlier? In particular, for example to prove that $$\gamma_1$$ and $$\gamma_2$$ have the same factorisation, do we already need that triple intersections are path-connected? It looks like we only need that pairwise intersections must be path-connected to prove this?

Almost all of the vertices belong to three rectangles; for example, in the figure with the numbered rectangles, there is a vertex at the intersection of rectangles 2, 3, and 6, and another vertex at the intersection of rectangles 3, 6, and 7. When we insert the paths $$\bar{g}_v g_v$$, we want those paths to lie in all three $$A_{ij}$$'s corresponding to the three rectangles (whenever there are three).

Let's focus on the vertex at rectangles 2, 3, and 6 (call it $$v_{236}$$). For convenience, let's also notate by $$A_r$$ the $$A_{ij}$$ into which $$F$$ maps $$R_r$$. When we move downward from $$v_{236}$$ along the edge between 2 and 3 (as we do for $$\gamma_2$$), we're producing a homotopy class in either $$\pi_1(A_2)$$ or $$\pi_1(A_3)$$. That means that the entire loop $$\bar{g}_{v_{236}} \cdot F|\gamma_{236 \to 23} \cdot g_{v_{23}}$$ which represents this class must lie in both $$A_2$$ and $$A_3$$, where $$\gamma_{236 \to 23}$$ is the path between rectangles 2 and 3. In particular, $$\bar{g}_{v_{236}}$$ must lie in both $$A_2$$ and $$A_3$$.

However, we could also move to the right from $$v_{236}$$ to $$v_{367}$$, as we do for $$\gamma_3$$ (before continuing rightward to $$v_{347}$$). But this loop, which lies in $$A_3$$ and $$A_6$$, also starts with $$\bar{g}_{v_{236}}$$! That means that $$\bar{g}_{v_{236}}$$ must lie in $$A_2 \cap A_3 \cap A_6$$. Thus, in order to produce the path $$g_{v_{236}}$$ from the basepoint to $$F(v_{236})$$, we need to assume that $$A_2 \cap A_3 \cap A_6$$ is path-connected.

• Why do we want those paths to lie in all three $A_{ij}'s$? For example, when going from $\gamma_2$ to $\gamma_3$ choosing a path in the common intersection of $2$ $A_{ij}'s$ seem to suffice?
– user661541
Dec 3, 2019 at 16:19
• I've edited my answer to better address this. Dec 3, 2019 at 16:44
• Thanks! In your answer, there seem to be two important vertices: $v_{236}$ and the vertex next to it $v_{367}$. But if we go from $\gamma_2$ to $\gamma_3$, the vertex $v_{367}$ does not play any role?
– user661541
Dec 3, 2019 at 16:55
• When Hatcher writes "Then we obtain a factorization of $[F|\gamma_r]$ by inserting the appropriate paths $\bar{g}_v g_v$ into $F|\gamma_r$ at successive vertices," I interpret this to mean that on the way from $v_{236}$ to $v_{347}$, you need to stop at $v_{367}$ and insert a $\bar{g}_v g_v$ here. If we didn't, we would run into trouble going from $\gamma_5$ to $\gamma_6$. Dec 3, 2019 at 16:58
• Ahah, I interpreted this in another way. Don't stop at $v_{367}$. Suppose we interpret it in my way. Do you think we need the triple intersections then at this stage of the proof?
– user661541
Dec 3, 2019 at 17:04