Check if this function injective or surjective $ f(x,y)=[x]_R \cap[y]_R$ Let $ R \subseteq \mathbb{N}\times\mathbb{N}$ be an equivalance relation. Function $ f:  \mathbb{N}\times\mathbb{N} \to \mathcal{P}(\mathbb{N})$ is describes as $ f(x,y)=[x]_R \cap[y]_R$
1)Check if it is injective.
2)Check if it is surrjective.
3)Find $f^{-1}(([3]_R))$
I can`t quite understand what $[x]_R$ notation means(the index of relation).Can some one explain what relation has to do here?
I think it is not injective beacuse we can change places of x and y.
 A: Let us see that $f$ is not injective, no matter what equivalence relation $R$ is.
Suppose that $R=\mathbb N\times \mathbb N$ (the largest equivalence relation on $\mathbb N$).
Then $[n]_R=\mathbb N$, for all $n$, and then $f(n,m)=\mathbb N$, for every $n,m$, and so it's not injective.
Now suppose that $R \neq \mathbb N \times \mathbb N$.
So there exist at least two equivalence classes of $R$ and one of them is not a singleton.
Let us say that $(x,y) \notin R$ and $(x,y') \notin R$, with $y\neq y'$, but $(y,y') \in R$ and thus, $[y]_R=[y']_R$.
Then
$$f(x,y) = [x]_R \cap [y]_R = [x]_R \cap [y']_R = f(x,y').$$
A similar argument can show it's not surjective.

Edit (I forgot the last part.)
Clearly,
$$f^{-1}([3]_R)= \{(x,y)\in\mathbb N^2 : f(x,y)=[3]_R\},$$
and that shouldn't be difficult to unfold.

Edit 2.
I just noticed that the OP claims to think $f$ not injective because

we can change places of $x$ and $y$.

That is indeed enough, since for $x\neq y$ we have $(x,y)\neq(y,x)$, but $f(x,y)=f(y,x)$.
So that makes a shorter proof that $f$ is not injective.

Edit 3. (OP ask for clarification.)
To show it's not surjective, again, suppose $R=\mathbb N \times \mathbb N$, so that $(n,m)\in R$ for all $m,n \in \mathbb N$.
Thus, for all $n$ we have $[n]_R=\mathbb N$, whence $\mathrm{im}(f)=\{\mathbb N\}$.
But $\mathbb N$ is one among (uncountably) infinite elements of $\mathcal P(\mathbb N)$, and so $f$ is not surjective.
On the other hand, if $R\neq \mathbb N \times \mathbb N$, then for all $n$, the equivalence class $[n]_R$ is a proper subset of $\mathbb N$, and so there exist no $x,y$ such that $[x]_R\cap[y]_R=\mathbb N$, and therefore, $\mathbb N \notin \mathrm{im}(f)$.
(Notice this would work even if we had a two-element set in place of $\mathbb N$.)
For the last part,
\begin{align}
f^{-1}([3]_R)
&= \left\{(x,y) \in \mathbb N^2 : f(x,y) = [3]_R\right\}\\
&= \left\{(x,y) \in \mathbb N^2 : [x]_R\cap[y]_R=[3]_R\right\}.
\end{align}
Now notice that two equivalence classes are either equal or disjoint, but $[3]_R \neq \varnothing$ because $3 \in [3]_R$, and so
$$[x]_R\cap[y]_R=[3]_R \iff [x]_R=[y]_R=[3]_R.$$
Hence
$$f^{-1}([3]_R)=\{(x,y) \in \mathbb N^2 : x,y\in[3]_R\}.$$
