Equivalence Relations and Mappings

Question

I have come across this statement in a book of Mathematical Physics. A map $\phi\colon X \to Y$ defines an equivalence relation R on the domain X by aRb if and only if ф(a)=ф(b). How to proceed with the properties of the equivalence relation and end up at ф(a)=ф(b).

Answer 1

You have misconstrued/misunderstood the role that the "if and only if" is playing in the statement.

The "if and only if" is part of the definition of $R$, not part of what you are trying to prove/the statement being given.

Specifically, definitions are almost always "if and only if" statements: we say X if and only if Y; that is, when Y, then we say X, when we say X, that means that Y.

For instance:

We say a subset $\mathbf{W}$ of a vector space $\mathbf{V}$ is a *subspace of $\mathbf{V}$ if and only if it is a vector space in its own right under the operations of $\mathbf{V}$ restricted to $\mathbf{W}$.

This means that: (i) if we say that $\mathbf{W}$ is a subspace of $\mathbf{V}$, that means that it is a subset that is a vector space under the restricted operations; and (ii) that $\mathbf{W}$ happens to be a subset of $\mathbf{V}$ that is a vector space under the restricted operations, then we will say $\mathbf{W}$ is a subspace.

You will sometimes see people who make definitions stating only "if". For example,

Two integers $m$ and $n$ are relatively prime if their greatest common divisor is $1$.

Formally, that says that if their gcd is $1$, then $m$ and $n$ are relatively prime. But it may seem like it leaves open the possibility that $m$ and $n$ are "relatively prime" under some other conditions too. That is, we are saying "A if B", which formally means $B\implies A$; this certainly means that if $B$ is true than $A$ is true, but it does not mean that if $A$ is true then $B$ must necessarily also be true.

So to avoid that possible reading, we say that $m$ and $n$ are relatively prime if and only if their greatest common divisor is $1$. This is a definition. It says that is the situation under which we will say they are relatively prime, and the only situation in which we will say so.

Similarly with the issue at hand. You are given two sets $X$ and $Y$, and a function $\phi\colon X\to Y$. We want to define a binary relation $R$ on $X$. We say:

$xRy$ if and only if $\phi(x)=\phi(y)$.

This means: when $\phi(x)=\phi(y)$, we will then have $xRy$; and if we have $xRy$, then it must be the case that $\phi(x)=\phi(y)$.

The statement to be proven then is just:

$R$ is an equivalence relation on $X$.

You seem to have misunderstood the statement as saying

The following are equivalent: 1. $R$ is an equivalence relation on $X$. 2. $\phi(x)=\phi(y)$.

Which of course would be difficult to establish, as it is nonsensical as written, because there is no connection between $R$ and $\phi$.

I hope this helps.

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At the risk of being confusing, here is something which is true:

Let $X$ be a set, and let $R$ be a relation on $X$. The following are equivalent:

1. There exists a set $Y$ and a function $\phi\colon X\to Y$ such that $xRy$ holds if and only if $\phi(x)=\phi(y)$.
2. $R$ is an equivalence relation.

The proof that 1 implies 2 is the statement you are proving here. The proof that 2 implies 1 has to do with the fact that an equivalence relation induces a partition on $X$ into equivalence classes, and that then one can define $Y$ to be the set of equivalence classes under $R$, denoted $X/R$, and let $\phi$ be the function that sends each $x\in X$ to its equivalence class in $Y$. But this is a different statement from the one you have on hand.