"Let $A$ and $B$ be sets and let $f : A \to B$ be a function. Define a relation on $B$ as follows. If $a, b \in B$, we say that $a\mathrel{R}b$ if and only if there exists some $c \in A$ such that $f(c) = a$ and $f(c) = b$. Prove that $R$ need not be an equivalence relation."
I'm having trouble understanding why this is not necessarily an equivalence relation. Based off the discussion in class about this problem, the professor stated that is not an equivalence relation because it fails to achieve the reflexivity criteria, but I do not see why.
If my memory serves me correctly, the counterexample given was,
Let $A = \{ 1 \}$ and $B = \{ 1,2 \}$ and $f(1) = 1$. Then $(2,2) \notin R$, so it is not reflexive and not an equivalence relation.
But this seems like it is not an equivalence relation simply because of the way we defined our sets and relation.
$\mathcal R$ is not necessarily an equivalence relation ON $B$ because $f:A\to B$ is not necessarily a surjection. For example, let $A=B=\Bbb R$ and consider the "zero function" $f:\Bbb R\to\Bbb R$ defined by $f(x)=0$. Clearly, $(1,1)\notin\mathcal R$ because there is no $c\in\Bbb R$ such that $f(c)=1$. Therefore, in this example, $\mathcal R$ is not an equivalence relation on $B$.