Let $f: \{1,2,3,4\} \to \{1,2,3,4\}$ be a function. We define the set of functions "$A = \{g: \{1,2,3,4\} \to \{1,2,3,4,5,6,7,8\}\}$" the relation $R$ given by
$gRh$ if and only if $g \circ f = h \circ f$
i)prove that $R$ is an equivalence relation on $A$
ii)assuming that $f$ is surjective, calculate the equivalence class of $g \in A$
OK. The first problem I have is that it already confuses me with that set "$A$" that has a function inside. I always work with $X = \{n\in N: n <2019\}$ for example ... I know what reflexivity, Symmetry, asymmetry and transitivity are ... but I honestly don't know how to face this problem. On the other hand I know that an equivalence relation means that it is reflexive, symmetric and transitive. That is, if it is $g\circ f = h\circ f$, does that mean that the function $g =$ the function $h$? If you could help me it would be great.
$A$ is a set of functions, in particular the set of all functions that map the numbers $\{1,2,3,4\}$ into the set $\{1,2,3,4,5,6,7,8\}$. For example, if $g(1) = 5, g(2)= 3, g(3) = 3, g(4) = 7$m then $g$ is an element of $A$.
So say you want to know if $R$ is reflexive. Is $gRg$ for every $g$, in other words. That is defined as $g \circ f = g \circ f$, so yes, it is reflexive.
If math is getting too abstract, make a concrete example. Suppose $f(1) = 1, f(2) = 1, f(3) = 3, f(4) = 3$