Proving two functions are equal

Question

Let $A$ and $B$ be subsets of the set $\Bbb Z$ for all integers, and let $\mathscr F$ denote the set of all functions $f:A\rightarrow B.$ Assume $A = \{1,2,3\}$ and $B=\{1,2,...,n\}$ where $n\ge 2$ is a fixed integer.

Define a relation $R$ on $\mathscr F$ by : for any $f,g \in \mathscr F$, $\;fRg\;$ if and only if $f - g$ is a constant function (i.e. there is a constant $c$ so that $f(x) - g(x) =c$ for all $x\in A$).

Let $f_1\in \mathscr F$ be defined by: $f_1(1)=2,f_1(2)=n,f_1(3)=1$. Suppose that $g \in \mathscr F$ is arbitrary so that $gRf_1$. Prove that $g=f_1$ and thus $[f_1]$ is just $ \{f_1\}$.

My work ( not much):

so $gRf_1$ means $g - f_1 = c$ $\Rightarrow g = f_1+c.$ To prove $g=f_1$ I need to prove $g\subseteq f_1$ and $f_1\subseteq g$. So for $g\subseteq f_1$ , let $x\in g$ and I need to prove $x\in f_1$.

So I am not sure where to go from here.

Answer 1

You seem to try equality of functions as one would with equality for sets. This is ok if one considers a function $A\to B$ ass a subset of $A\times B$. Here, however, it is much easier to argue with function values.

Assume $fRg$, so there is $c\in\mathbb Z$ with $f(x)-g(x)=c$ for all $x\in A$. Since $1\le g(x)\le n$ for all $x$, we find $$1\le g(3)=f(3)-c=1-c\implies c\le0$$ and $$n\ge g(2)=f(2)-c=n-c\implies c\ge0,$$ hence $c=0$, i.e. $f=g$.

Answer 2

Another way to approach this is if we prove that $c = 0$.

We know the following. $$g(1) = 2+c$$ $$g(2) = n+c$$ $$g(3) = 1+c$$

$g(2) \le n \Rightarrow $ $n+c \le n$ $\Rightarrow$ $c\le 0$.

Thus, $g(3) \le 1$ and since $1$ is the smallest member of $B$, $g(3) = 1$.

Therefore, c = 0 and $g = f_1$.