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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – user661541
    Dec 3, 2019 at 16:19
  • $\begingroup$ I've edited my answer to better address this. $\endgroup$ Dec 3, 2019 at 16:44
  • $\begingroup$ 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? $\endgroup$
    – user661541
    Dec 3, 2019 at 16:55
  • 1
    $\begingroup$ 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 $\gamma_6$. $\endgroup$ Dec 3, 2019 at 16:58
  • $\begingroup$ 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? $\endgroup$
    – user661541
    Dec 3, 2019 at 17:04

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