Hypotheses

If there exists a bijective function from ${\displaystyle A}$ to ${\displaystyle B}$, then ${\displaystyle \left|A\right|=\left|B\right|}$.

Given ${\displaystyle f:A\to B}$ where ${\displaystyle A=B}$ and ${\displaystyle A}$ and ${\displaystyle B}$ are finite, any surjective or injective function must be bijective.

Given ${\displaystyle f:A\to A}$ where ${\displaystyle A}$ is infinite, there exist functions that are injective but not surjective and vice versa.

Proof: Assume by way of contradiction that ${\displaystyle B=\varnothing }$ but ${\displaystyle A\ne \varnothing }$. Then there is an element ${\displaystyle u\in A}$. By the the definition of a function, there exists an element ${\displaystyle v\in B}$ such that ${\displaystyle (u,v)\in A\times B}$. But ${\displaystyle B=\varnothing }$ so there can be no such element. Therefore, if ${\displaystyle B=\varnothing }$, then ${\displaystyle A=\varnothing }$.

Proof: This theorem is really two theorems in one, so we'll address each separately.

1. Assume by way of contradiction that ${\displaystyle A=\varnothing }$ but ${\displaystyle f\ne \varnothing }$. Then by the definition of a function, there is an element ${\displaystyle (u,v)\in A\times B}$, but because ${\displaystyle A=\varnothing }$, this cannot be the case. Therefore, if ${\displaystyle A=\varnothing }$, then ${\displaystyle B=\varnothing }$.
2. Assume by way of contradiction that ${\displaystyle A=\varnothing }$ but ${\displaystyle Im\left(f\right)\ne \varnothing }$. Then, there exists an element ${\displaystyle v}$ in the image set. By the definition of image set, there exists an element ${\displaystyle u\in A}$ such that ${\displaystyle f\left(u\right)=v}$. But since ${\displaystyle A=\varnothing }$, that cannot be the case. Therefore, if ${\displaystyle A=\varnothing }$, then ${\displaystyle Im\left(f\right)=\varnothing }$.

The number of distinct functions from a finite domain ${\displaystyle A}$ to a finite codomain ${\displaystyle B}$ is given by ${\displaystyle {\left|B\right|}^{\left|A\right|}}$. This is the number of distinct lists of length ${\displaystyle \left|A\right|}$ that can formed using elements from ${\displaystyle B}$.

The number of distinct injective functions from a finite domain ${\displaystyle A}$ to a finite domain ${\displaystyle B}$ is given by ${\displaystyle \{\begin{array}{ll}0& if\left|A\right|>\left|B\right|\\ \frac{\left|B\right|!}{(\left|B\right|-\left|A\right|)!}& otherwise\end{array}\}}$. This is equal to the number of permutations of ${\displaystyle B}$ containing ${\displaystyle \left|A\right|}$ elements.

The number of distinct surjective functions from a finite domain ${\displaystyle A}$ to a finite domain ${\displaystyle B}$ is given by ${\displaystyle \{\begin{array}{ll}0& if\left|A\right|<\left|B\right|\\ {\left|B\right|}^{\left|A\right|-\left|B\right|}\times \left|B\right|!& otherwise\end{array}\}}$. This is equal to the number of permutations of ${\displaystyle B}$ times the number of unique lists of length ${\displaystyle \left|A\right|-\left|B\right|}$ that can be formed using elements from ${\displaystyle B}$.

The number of distinct injective functions from a finite domain ${\displaystyle A}$ to a finite domain ${\displaystyle B}$ is given by ${\displaystyle \{\begin{array}{ll}0& if\left|A\right|\ne \left|B\right|\\ \left|B\right|!& otherwise\end{array}\}}$. This is equal to the number of permutations of ${\displaystyle B}$.