Mathematical induction proving f is not one to one using n

Question

I am unsure about how to go about this proof which is supposed to prove the pigeonhole principle using mathematical induction on n by proving f is not one-to-one.

"For every n≥1, every function f: A→{1,2, . . . , n}where|A|> n,f is not one-to-one."

Answer 1

Well, Okay... then...

Base case:

Every function $f: A \to \{1\}$ where $|A| > 1$ is not one to one.

Pf: Since $|A| > 1$ then $|A|\ge 2$ so there exist $a,b \in A$ so that $a \ne b$.

But $f(a) = 1; f(b) = 1$. So $f$ is not one to one.

Induction step:

If every function $f: A \to \{1..... k\}$ where $|A| > k$ is not one to one, then every function $g: B\ to \{1....k+1\}$ where $|B| > k+1$ isnt one-to one.

Hint: If $g: B \to \{1......k+1\}$ is one to one, and for $a\in B$ then if $g(a) = j$ then for all $b \in B$ so that $g(b) \ne j$.

So let $h: B\setminus\{a\} \to \{1..... k+1\}\setminus\{j\}$ where $h(b) = g(b)$ for all $b \in B\setminus\{a\}$ is well defined.

Questions: Is $h$ one to one? (Hint: the answer is yes.) Is it possible for $h$ to be one to one? (Hint: the answer is no.)

Hint 4: $\{1..... k+1\}\setminus\{j\}$ is isomorphic to $\{1..... k\}$. And $|B\setminus\{a\}| = |B| -1 > k+1 -1 = k$.