Determine whether the following relations are injective and/or subjective function. Find universe of the functions if they exist. i. A= v,w,x,y,z, B=1,2,3,4,5 R= (v,z),(w,1), (x,3),(y,5) ii. A = 1,2,3,4,5 B=1,2,3,4,5 R = (1,2),(2,3),(3,4),(4,5),(5,1)

We know that a function is called injective if the values of the function are equal if and only when the arguments are equal $(x;z), (y;z) \in R\iff (x=y)$ .

The function is called subjective if For each element of the set B, there is its inverse image with respect to the function: $\forall a\in B\implies \exists x\in A : (x;a)\in R$ .

i.In this case $R$ is not a function acting from set $A$ t the set $B$ , becuse in $(v,z)$ $z\notin B$ . It is not a function at all, s it can't be injective and/or subjective.

ii. In this case for all values of the function $E(R)=\{1,2,3,4,5\}$ there exist different arguments from the set $A$ . So the function $R$ is injective. $E(R)=\{1,2,3,4,5\}$ equals $B$ , so the function $R$ is subjective. The universe of the function is $A\times B=\{(1;1), (1;2),(1;3),(1;4),(1;5),(2;1),(2;2)$  $(2;3),(2;4),(2;5),(3;1),(3;2),(3;3),(3;4),(3;5),$ $(4;1),(4;2),(4;3),(4;4),(4;5), (5;1),(5;2),(5;3),(5;4),(5;5)\}$

The inverse function to the function $R$ is $R^{-1}=\{(1;5), (2;1),(3;2),(4;3), (5;4)\}$.