Why do we use pigeonhole principle to prove equality? I often found that to prove $2$ variables are equal,  pigeonhole principle is used. For example see the following text:

Let $F$ be any family of type $t_{i, j}$ ... and so the family $F$ is mapped into a family $F'$ of edges of type
$t_{r, s}, r\geq i, s\geq j$. By the pigeonhole principle,  $r = i$ and
$s = j$, from which the lemma follows (pages 120, 121, 122).

We can directly say $r = i$ and $s = j$, why $r > i$ and $s> j$ was assumed and  pigeonhole principle is used?
Added:
If $F$ is a family of edges of type $t_{i,j}$, then ($F)_k$ denotes the set of $j$ vertices of $V_k$
incident to the family, and $(F)_{k+1}$ denotes the set of $i$ vertices in $V_{k+1}$ incident to the family.
If you look at the definition on page 121 (second last paragraph):$F$ is a family of edges of type $t_{i,j}$, then ($F)_k$ denotes the set of $j$ vertices of $V_k$ incident to the family, and $(F)_{k+1}$ denotes the set of $i$ vertices in $V_{k+1}$ incident to the family. That is why if a permutation fixes  $(F)_k$ it also fixes  $(F)_{k+1}$ (AND VICE VERSA, to my understanding), because of the adjacency.
Similarly, if a  permutation $\pi$ moves $(F)_k$ consisting $j$ vertices to another $(F')_k$ (which also consists $j$ vertices), then because of the adjacency defined,   $\pi$ moves the incident $(F)_k$ of $i$ vertices, to another $(F')_k$  of $i$ vertices.
In this situation, why would you consider the case $j = |(F)_k| < |(F')_k| = s$ and $i = |(F)_{k+1}| <  |(F')_{k+1}| = r$, instead of directly equating?
 A: We are trying to prove that $\alpha$ maps a family $F$ of type $t_{i,j}$ to a family of the same type - so no, we cannot assume that $F'$ has the same type.
What we know is that $\alpha$ maps $(F)_k$ into $(F')_k$, so $j = |(F)_k| \le |(F')_k| = s$. Similarly, since $\alpha$ maps $(F)_{k+1}$ into $(F')_{k+1}$, we have $i = |(F)_{k+1}| \le |(F')_{k+1}| = r$. This is what tells us that $F'$ has some type $t_{r,s}$ with $i \le r$ and $j \le s$. (But as of this moment, we haven't ruled out the possibility that $(F')_k$ has other vertices in addition to the $j$ that are the image of $(F)_k$ under $\alpha$; this is why we cannot immediately say that $j=s$. Similarly for $i$ and $r$.)
The argument to show $i=r$ and $j=s$ is not pigeonhole, exactly, though it has the same flavor. Let $U$ be the union of all families of types $t_{r,s}$ with $r+s > i+j$. We've shown that $\alpha$ maps every family contained in $U$ into some other family contained in $U$. So actually, $\alpha$ must simply permute $U$ somehow.
Therefore $F'$ cannot have a type $t_{r,s}$ with $r+s > i+j$: otherwise, it would be in $U$, so its preimage would be in $U$. Since we know $r+s \ge i+j$, we conclude $r+s = i+j$, so $r=i$ and $s=j$.
