Lemma 1   Let $A,B$ be two finite sets of the same cardinality. If $f\colon A \to B$ is an injective function then $f$ is bijective.

Proof. In order to prove the lemma, it suffices to show that if $f$ is an injection then the cardinality of $f(A)$ and $A$ are equal. We prove this by induction on $n={card}(A)$ The case $n=1$ is trivial. Assume that the lemma is true for sets of cardinality $n$ and let $A$ be a set of cardinality $n+1$ Let $a\in A$ so that $A_1=A-\{a\}$ has cardinality $n$ Thus, $f(A_1)$ has cardinality $n$ by the induction hypothesis. Moreover, $f(a)\notin f(A_1)$ because $a\notin A_1$ and $f$ is injective. Therefore: $$f(A)=f(\{a\}\cup A_1)=\{f(a)\}\cup f(A_1)$$ and the set $\{f(a)\}\cup f(A_1)$ has cardinality $1+n$ as desired.