I have been reading Hatcher's proof of Van Kampen's theorem on page 45, and there is an argument that I don't understand:
$X$ is the union of path-connected open sets $A_\alpha$ each containing the basepoint $x_0\in X$ and each intersection $A_\alpha\cap A_\beta\cap A_\gamma$ is path-connected.
Let $[f_1]\cdots[f_k]$ and $[f'_1]\cdots[f'_ℓ]$ be two factorizations of $[f]\in\pi_1(X)$. The composed paths $f_1\cdots f_k$ and $f'_1\cdots f'_ℓ$ are then homotopic, so let $F:I\times I\to X$ be a homotopy from $f_1\cdots f_k$ to $f'_1\cdots f'_ℓ$. There exist partitions $0 = s_0 < s_1 < \cdots < s_m = 1$ and $0 = t_0 < t_1 <\cdots < t_n = 1$ such that each rectangle $R_{ij} = [s_{i−1}, s_i]×[t_{j−1}, t_j]$ is mapped by $F$ into a single $A_\alpha$, which we label $A_{ij}$. These partitions may be obtained by covering $I×I$ by finitely many rectangles $[a, b]×[c, d]$ each mapping to a single $A_\alpha$, using a compactness argument, then partitioning $I×I$ by the union of all the horizontal and vertical lines containing edges of these rectangles. We may assume the $s$ partition subdivides the partitions giving the products $f_1\cdots f_k$ and $f'_1\cdots f'_ℓ$. Since $F$ maps a neighborhood of $R_{ij}$ to $A_{ij}$, we may perturb the vertical sides of the rectangles $R_{ij}$ so that each point of $I\times I$ lies in at most three $R_{ij}$’s. We may assume there are at least three rows of rectangles, so we can do this perturbation just on the rectangles in the intermediate rows, leaving the top and bottom rows unchanged. Let us relabel the new rectangles $R_1, R_2, \cdots , R_{mn}$, ordering them as in the figure.
Question: I don't know where the "three" comes from. But intuitively, suppose a point $F(s,t)$ is in the intersection of four $A_\alpha$'s, then how do we perturb the vertical sides of the rectangles in $I\times I$ so that $F(s,t)$ lies in three of the $A_{\alpha}$'s? I can't see how this is possible.
A point in $I\times I$ that lies in the intersection of four rectangles is basically the coincident vertex of these four.Then we "perturb the vertical sides" of some of them so that the point lies in at most three $R_{ij}$'s and for these four rectangles,they have no vertices coincide.And since $F$ maps a neighborhood of $R_{ij}$ to $A_{ij}$,we can make a sufficiently small perturbation so that the resulting rectangle is contained in the very neighborhood for each one.