How would you deal with equivalence relation and equivalence classes with functions??

Question

Suppose a function $f:A→B$ is given. Define a relation ~ on $A$ as follows:

$a_1$~$a_2$ ⟺ $f(a_1)$ = $f(a_2)$

Since ~ is an equivalence relation, it induces a partition of $A$ into equivalence classes. Describe these equivalence classes in each of the following cases. ($R$ is the set of real numbers).

(a) $A = B = R, f(x)=x^2$

(b) $A = B = R, f(x)=|x|$

(c) $A = R×R , B = R, f(x,y)=x+y$

my approach:

(a): the equivalence class contains no elements because the square of different numbers is different.

(b) : the equivalence class contains the set of real numbers.

(c): don't know about c

Is my approach anywhere near to correct? I am not so sure about the answers. please help!

Answer 1

You are starting your approaches with "the equivalence class contains …" as if there was only one equivalence class in each example. This is wrong. For each $x\in A$ there is an equivalence class $[x]=\{y\in A\mid x\sim y\}$ and this is never empty as it always contains $x$ itself.

In the first example consider $x=2$. We have $x\sim y$ if and only if $x^2=y^2$ so $y^2=4$, which has the solutions $-2$ and $2$. This means that $[2]=\{-2,2\}=[-2]$. The same argument shows that $[x]=\{-x,x\}$ for all $x\in\mathbb R$ with $x\neq 0$. Finally, since $y^2=0^2$ has the unique solution $y=0$, we have $[0]=\{0\}$ as the only equivalence class with just one element, while all other classes have two elements.

With this input I'll leave it to you to try the other examples again.

Answer 2

Your approach in the first question is wrong, because the square of $x$ and $-x$ are the same,

though $x$ and $-x$ are different numbers unless $x=0$.

Likewise, $\lvert x \rvert=\lvert -x \rvert$, so again the equivalence classes are $\{x,-x\}$ where $x\in\mathbb R$.

For the last one, $(x_1,y_1)\sim(x_2,y_2)\iff x_1+y_1=x_2+y_2.$ Say $x_1+y_1=c$.

Then the points equivalent to $(x_1,y_1)$ lie on the line $y=-x+c$.