What's the term for my function's aspect of equivalence structure preservation? $A, B$ are two sets, $\equiv$ is an binary equivalence relation on $A$, and $f  : A \rightarrow B$ is a function. If $x \not \equiv y \Rightarrow f(x) \neq f(y)$, what does that make $f$?
This answer suggests that

"Homomorphism," roughly speaking, refers to a map between sets equipped with some kind of structure that preserves that structure. 

and my equivalence relation is a structure $A$ is equipped with, and $f$ preserves it - after a fashion. But there's no group, or vector space, which would let me say $f(x \circ y) = f(x) \circ f(y)$. 
The Wikipedia page for Equivalence relations says:

if $x_1 \sim x_2$ implies $f(x_1) = f(x_2)$ then $f$ is said to be a morphism for $\sim$

But I have the converse of that.
So, what would I call a function like $f$?

A working mathematician would probably say something like "the induced map $\hat{f}:A/\!\!\equiv\; \rightarrow B$ is injective" for this condition. Something like that.
Edit. As the OP points out, what I just wrote is incorrect; no such map is necessarily induced, because we're not assuming that $f$ turns $\equiv$ into $=$. Given this, I don't think there's a standard way of writing the condition of interest. But you could still say something like: "the induced relation $\hat{f} : A/\!\!\equiv\,\, \rightarrow B$ is injective.
In more detail: given an equivalence relation $\equiv$ on a set $A$, write $[\equiv]: A \rightarrow A/\!\!\equiv$ for the corresponding projection. We get a corresponding relation $[\equiv]^\dagger : A/\!\!\equiv \,\,\rightarrow A$. The condition you're looking for is that the relation $f \circ [\equiv]^\dagger : A/\!\!\equiv\,\, \rightarrow B$ is injective, or in other words that the relation $[\equiv]\circ f^\dagger$ is deterministic.
