Prove ##f^{-1}(G\cap H)=f^{-1}(G)\cap f^{-1}(H)##

[Potatochip911]

Homework Statement

So as the title says I'm supposed to prove $f^{-1}(G\cap H)=f^{-1}(G)\cap f^{-1}(H)$

Homework Equations

3. The Attempt at a Solution [/B]
I've been doing a lot of proofs for my class lately and just wanted to make sure that I'm at least doing one of them correctly so I would appreciate it if someone could tell me whether or not this is correct.

1) Let $x\in f^{-1}(G\cap H)$ then $\exists y\in (G\cap H)$ such that $y=f(x)$ so $y\in G$ and $y\in H$ therefore $x\in f^{-1}(G)$ and $x\in f^{-1}(H)$ so $x\in f^{-1}(G)\cap f^{-1}(H)$ which implies $f^{-1}(G\cap H)\subseteq f^{-1}(G)\cap f^{-1}(H)$

2) Let $x\in f^{-1}(G)\cap f^{-1}(H)$, so $x\in f^{-1}(G)$ and $x\in f^{-1}(H)$. $\exists a\in G$ such that $a=f(x)$ and $\exists b\in H$ such that $b=f(x)$, since $a=b$; $a\in G\cap H$ so $x\in f^{-1}(G\cap H)$ which implies $f^{-1}(G)\cap f^{-1}(H)\subseteq f^{-1}(G\cap H)$

From 1) and 2) we have that $f^{-1}(G\cap H)=f^{-1}(G)\cap f^{-1}(H)$

 

[tommyxu3]

It seems no problems that $A\subseteq B$ and $B\subseteq A$ imply $A=B.$

 

[Fredrik]

It looks good, but it can be simplified a bit. For the first part, I would just note that the following implications hold for all x.
$$x\in f^{-1}(G\cap H)\ \Rightarrow\ f(x)\in G\cap H\ \Rightarrow\ \begin{cases}f(x)\in G\\ f(x)\in H\end{cases}\ \Rightarrow\ \begin{cases}x\in f^{-1}(G)\\ x\in f^{-1}(H)\end{cases}\ \Rightarrow\ x\in f^{-1}(G)\cap f^{-1}(H).
$$ For the second part, I would just stare at these implications until I have convinced myself that they all hold in the opposite direction as well. That strategy doesn't always work, but it does in this case.