If $P\subset X$ and $Q\subset Y$,define $f(P)=\{f(x):x\in P\}$ and $f^{-1}(Q)=\{x:f(x)\in Q\}$. Then the true statement(s) is/are:

Question

Let $f:X\to Y$ be a function. If $P\subset X$ and $Q\subset Y$,define $f(P)=\{f(x):x\in P\}$ and $f^{-1}(Q)=\{x:f(x)\in Q\}$. Then the true statement(s) is/are

1. $f(f^{-1}(Q))=Q$

2. $f^{-1}(Q)\cup f^{-1}(R)=f^{-1}(Q\cup R)$

3. $f^{-1}(f(P))=P$

4. $f^{-1}(Q)\cap f^{-1}(R)=f^{-1}(Q\cap R) $

The correct answers are given to be 2. and 4.

But why can't 1. and 3. be the correct answers and I am not able to come up with counter examples for 2. and 4. . Is there a proper way to prove these statements

Answer 1

It's always good to try simple counterexamples.

For 1. consider $f: \{1,2,3\} \to \{1,2,3\}$ given by $f(1)=f(2)=f(3)=1$. Then $f^{-1}(\{1,2\}) = \{1,2,3\}$ but $f(\{1,2,3\}) = \{1\}$. Now try and prove yourself that 1. is true iff $f$ is surjective.

To prove statements like 2. and 4. you need to prove both inclusions separately. Start with an element in one set and show that it belongs to the other.

I'll leave 3. to you as it's quite similar to 1. Just like 1. you can also generalize it.