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.
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
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 $$f^{-1}([3]_R)=\{(x,y) \in \mathbb N^2 : x,y\in[3]_R\}.$$