Is the relation $a\mathrel R b \iff f(a) \equiv f(b)$ an equivalence relation? Suppose that I have a relation $R$ of the form $a\mathrel R b \iff f(a) \equiv f(b)$, where $\equiv$ is an equivalence relation.
In general, is $R$ also an equivalence relation? If not, what are the sufficient conditions on $f$ such that $R$ is an equivalence relation?
It seems intuitive that so long as $f$ is some deterministic function, then the properties of $\equiv$ are inherited by $R$, making it also an equivalence relation.
 A: If $X$ and $Y$ are sets and $f:X \to Y$ is a function, then $R$ is an equivalence relation on $X$ if "$\equiv$" is an equivalence relation on $Y$.  We need merely check the axioms:
I.)  Reflexitivity:  for any $x \in X$, we have $f(x) \equiv f(x)$, so $x\ R\ x$;
II.)  Symmetry:  for $x, y \in X$, we have $f(x) \equiv f(y) \Leftrightarrow f(y) \equiv f(x)$;  thus $x\ R\ y \Leftrightarrow y\ R\ x$;
III.)  Transitivity:  for any $x, y, z \in Z$, $x\ R\ y \Leftrightarrow f(x) \equiv f(y)$ and $y\ R\ z \Leftrightarrow f(y) \equiv f(z)$; but $f(x) \equiv f(y)$ and $f(y) \equiv f(z)$ imply $f(x) \equiv f(z)$ so $x\ R\ z$; $R$ is indeed transitive.
It follows from (I)-(III) that $R$ is indeed an equivalence relation on $X$.
Addenda, Tuesday 26 August 2014 1:41 PM PST:  I wanted to add a few further remarks, but it was late when I answered, so I posted what I had and waited for a new day.
This being said:
IV.)  We have seen that $R$ defined as above is an equivalence relation, "inheriting" this property from "$\equiv$"; in the event that $f:X \to Y$ is surjective, the "inheritance works both ways, that is, "$\equiv$" is an equivalence relation if $R$ is.  This is easily seen using the surjectivity of $f$; for example, to show transitivity we observe that if $f(x) \equiv f(y)$ and $f(y) \equiv f(z)$, then $x\ R\ y$ and $y\ R\ z$, so $x\ R\ z$ whence $f(x) \equiv f(z)$.  The reflexitivity and symmetry  of $\equiv$ similarly follow;
V.)  There is an alternative approach to establishing that $R$ is an equivalence relation via examination of the equivalence classes of $\equiv$; as is well-known, $\equiv$ partitions $Y$ into a collection of disjoint subsets $Y_i$ where $i \in \mathcal O$ some set of indices.  Then using the well-known and elementary propeties $f^{-1}(\bigcup_{i \in \mathcal Q} Y_i) = \bigcup_{i \in \mathcal Q}f^{-1}(Y_i)$ and $f^{-1}(\bigcap_{i \in \mathcal Q} Y_i) = \bigcap_{i \in \mathcal Q} f^{-1}(Y_i)$ for $\mathcal Q \subset \mathcal O$ one may easily see that the partition of $Y$ induced by $\equiv$ yields a partition $f^{-1}(Y_i)$ of $X$; since equivalence relations and partitions determine each other uniquely, we see that $R$ is such a relation on $X$, corresponding the $f^{-1}(Y_i)$; I leave the remaining details to my readership.  End of Addenda.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
