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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 $[F|\gamma_{r+1}]$.

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  • $\begingroup$ Hi. Thanks for the answer. Could you explain a little more why $F\vert \gamma_r$ is homotopic to $F\vert \gamma_{r+1}$? $\endgroup$
    – user661541
    Commented Dec 3, 2019 at 10:26
  • $\begingroup$ Of course! I've edited the first paragraph of my answer to include a bit more about the homotopy. $\endgroup$ Commented Dec 3, 2019 at 10:43
  • $\begingroup$ 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 $h_1*h_2$ and $g_1*g_2$ 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 $R_{ij}$. $\endgroup$
    – user661541
    Commented Dec 3, 2019 at 10:51
  • $\begingroup$ Lifting this homotopy through $F$, we get that $F\vert \gamma_1$ and $F\vert \gamma_2$ are homotopic. Is that idea somewhat correct? $\endgroup$
    – user661541
    Commented Dec 3, 2019 at 10:51
  • $\begingroup$ 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 $F|\gamma_2$. $\endgroup$ Commented Dec 3, 2019 at 11:54

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