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In the proof of Van Kampen's theorem in Hatcher's book, which is theorem 1.20 on p43 , we read in its proof on p45 (see here: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf)

If anybody wants, I can make a screenshot and post it here.

Furthermore, the factorizations associated to successive paths $γ_r$ and $γ_{r+1}$ are equivalent since pushing $γ_r$ across $R_{r+1}$ to $γ_{r+1}$ changes $F \vertγ_r$ to $F \vert γ_{r+1}$ by a homotopy within the $A_{ij}$ corresponding to $R_{r+1}$ , and we can choose this $A_{ij}$ for all the segments of $γ_r$ and $γ_{r+1}$ in $R_{r+1}$.

What does one mean with 'pushing $\gamma_r$ across $R_{r+1}$'? Is this a homotopy? Can someone explain what is going on in this paragraph?

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Yes, "pushing $\gamma_r$ across $R_{r+1}$" means using a homotopy; $F|\gamma_r$ is homotopic to $F|\gamma_{r+1}$, since the restriction of $F$ to $R_{r+1}$ provides a homotopy between them via the square lemma (or a slight variation of the square lemma which allows for non-square rectangles). But there's more we can say; the factorization of $[F|\gamma_r]$ is equivalent to the factorization of $[F|\gamma_{r+1}]$.

Each of these two factorizations might contain a product of multiple terms corresponding to the edges around $R_{r+1}$ since, for most of the rectangles $R_i$, you'll hit about three vertices while traveling from one corner of the rectangle to the other (either way around the rectangle). However, all of those terms can be replaced with terms in $\pi_1(A_{ij})$, where $A_{ij}$ is the open set into which $F$ maps $R_{r+1}$; the last bullet point on page 44 says that replacing these terms gives new factorizations which are equivalent to the old ones. Then we can combine all the terms going clockwise around $R_{r+1}$ into a single term (using the second-to-last bullet point on page 44), and we can also combine all the terms going counterclockwise around $R_{r+1}$ into a single term; this gives equivalent factorizations. Now these two terms in $\pi_1(A_{ij})$ are equal, since they are the homotopy classes of two loops in $A_{ij}$ which are homotopic through a homotopy in $A_{ij}$; thus the factorization of $[F|\gamma_r]$ is equivalent to the factorization of $[F|\gamma_{r+1}]$.

Justin Barhite
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  • Hi. Thanks for the answer. Could you explain a little more why $F\vert \gamma_r$ is homotopic to $F\vert \gamma_{r+1}$? –  Dec 03 '19 at 10:26
  • Of course! I've edited the first paragraph of my answer to include a bit more about the homotopy. – Justin Barhite Dec 03 '19 at 10:43
  • The way I view it is: We can view for example $\gamma_1$ as the concatenation $f_1 * h_1 * h_2 * f_2$ (assuming three horizontal rectangles in the subdivision) and $\gamma_2$ as the subdivision $f_1 * g_1 * g_2 * f_2$ where these $f_i's, g_i's$ (not the same $f_i$'s as in the proof) are the paths along the segments. The path $h_1h_2$ and $g_1g_2$ have same begin-and end points and lie in the same square $R_{ij}$ and the square $R_{ij}$ is simply connected, so they are homotopic to each other in the square $R_{ij}$. –  Dec 03 '19 at 10:51
  • Lifting this homotopy through $F$, we get that $F\vert \gamma_1$ and $F\vert \gamma_2$ are homotopic. Is that idea somewhat correct? –  Dec 03 '19 at 10:51
  • Yes, exactly. The two paths around $R_{ij}$ are homotopic in $R_{ij}$, so composing with $F$ gives a homotopy from $F|\gamma_1$ to $F|\gamma_2$. – Justin Barhite Dec 03 '19 at 11:54
  • Thanks a lot! I think this is the hardest proof I did so far haha. It is very technical. –  Dec 03 '19 at 12:05
  • One last follow up question: We get a factorisation for $\gamma_1$ and a factorisation for $\gamma_2$. Do we already need that triple intersections are path connected here to see that these two factorisations are equivalent? My proof only seems to use that double intersections are path connected. I think one needs the triple intersections only to make sure $\gamma_0$ and $f_1* \dots * f_k$ have equivalent factorisations (and similarly for $\gamma_{mn}$ and $f_1'* \dots * f_l'$). –  Dec 03 '19 at 15:35
  • Does that make any sense? Or do you already need it earlier that triple intersections are pathconnected? –  Dec 03 '19 at 15:45
  • I asked a separate question here: https://math.stackexchange.com/questions/3461509/where-do-we-use-path-connectedness-in-the-proof-of-van-kampens-theorem –  Dec 03 '19 at 15:58