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proof of pigeonhole principle
If $|S|=0$ , then $S=\emptyset$ , and $f \colon \emptyset \to T$ can only be surjective if $T=\emptyset$ .
Assume the statement holds for any set $S$ with $|S|=n$ . Let $|S|=n+1$ . Let $x_1, \dots , x_{n+1} \in S$ with $S=\{x_1, \dots , x_{n+1}\}$ . Let $R=S \setminus \{x_{n+1}\}$ . Then $|R|=n$ .
Define $g \colon R \to T \setminus \{f(x_{n+1})\}$ by $g(x)=f(x)$ . Since $R \subset S$ , $f(x) \in T$ for all $x \in R$ . Thus, to show that $g$ is well-defined, it only needs to be verified that $f(x) \neq f(x_{n+1})$ for all $x \in R$ . This follows immediately from the facts that $x_{n+1} \notin R$ and $f$ is injective. Therefore, $g$ is well-defined.
Now it need to be proven that $g$ is a bijection. The fact that $g$ is injective follows immediately from the fact that $f$ is injective. To verify that $g$ is surjective, let $y \in T \setminus \{f(x_{n+1})\}$ . Since $f$ is surjective, there exists $x \in S$ with $f(x)=y$ . Since $f(x)=y \neq f(x_{n+1})$ and $f$ is injective, $x \neq x_{n+1}$ . Thus, $x \in R$ . Hence, $g(x)=f(x)=y$ . It follows that $g$ is a bijection.
By the induction hypothesis, $|R|=|T \setminus \{f(x_{n+1})\}|$ . Thus, $n=|R|=|T \setminus \{f(x_{n+1})\}|=|T|-1$ . Therefore, $|T|=n+1=|S|$ . 