Let $A$ and $B$ be two finite sets with $\vert A \vert = \vert B \vert$ and $f$ be a function from $A$ to $B$. show that if $f$ is an injective (one-to-one) function then it is also a surjective function.
Answer: i don't know if this assumption right or wrong
By contradiction that function $f$ one to one $\Rightarrow$ onto we assume that $f$ is not onto function $\forall a \in A∶f(a)\neq b$
$\vert A\vert=\vert B\vert$,then there exist two different elements in $A$ such that have same image one to one function then for $\exists x \in A ,\exists y \in B ,f(x)=f(y)\land (x\neq y)$.
However Since $f$ is one to one function,$f(x)=f(y)$ implies $x=y$ $\exists x \in A ,\exists y\in B ,f(x)=f(y)\land (x\neq y)$.
by negation law we obtain that last statement is F(false)and we obtain contradiction thus $f$ is onto since it's one to one.
Since $A$ and $B$ have the same number of elements, then an injective map from $A$ to $B$ exists. Injective means that two different elements $x,y \in A$ map to two different elements $f(x),f(y) \in B$. So every element of $A$ maps to exactly one element of $B$ therefore the map is also surjective since all elements of $B$ are mapped to some element in $A$.