$|A|=|B|=n\in N$
Prove/Disprove that if $f:A \rightarrow B$ is 1-1 then $f:A \rightarrow B$ is onto.
The answer I saw used the Pigeonhole Principle, I am trying to prove it without the principle. But I do not manage to prove that if cardinality is a finite number then all 1-1 functions must be onto too.
I mean that in the infinite case when two sets are equinumerous a bijection is exist, but in the finite case all 1-1 functions are bijection
On
We have to prove that an injective $f: \>[n]\to[n]$ is automatically bijective.
This is obviously true for $n=1$. Therefore assume that it is true as stated for some $n\geq1$ and consider an injective $g:\>[n+1]\to[n+1]$.
(a) When $g(n+1)=n+1$ then the restriction $f:=g\restriction[n]$ is an injective map from $[n]$ to $[n]$, and so is bijective in virtue of the induction hypothesis. It follows that $g$ is bijective.
(b) When $g(n+1)=a\leq n$ denote by $T:\>[n+1]\to[n+1]$ the transposition $a\leftrightarrow n+1$. Then $h:=T\circ g:\>[n+1]\to[n+1]$ is an injective map with $h(n+1)=n+1$ and so is bijective according to (a). As $T$ is clearly bijective it follows that $g=T\circ h$ is bijective.
Suppose $\exists$ an element $b \in B$ which is not mapped to an element in $A$. We know that $f$ is one-to-one. That is $f$ maps $n$ elements in $A$ to $n$ elements in $B$. Including $b$ the cardinality of $B$ is $(n + 1)$ leading to a contradiction.
And one more thing. Had to add this. If two sets have the same cardinality a bijection between them exists - by definition regardless of whether the range/domain is finite or not. But you've been given the function and are asked to prove that it is a bijection. You cannot use that fact in this particular question.