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

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data
    So as the title says I'm supposed to prove ##f^{-1}(G\cap H)=f^{-1}(G)\cap f^{-1}(H)##

    2. Relevant equations
    3. The attempt at a solution

    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)##
     
  2. jcsd
  3. Oct 3, 2015 #2
    It seems no problems that ##A\subseteq B## and ##B\subseteq A## imply ##A=B.##
     
  4. Oct 3, 2015 #3

    Fredrik

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    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.
     
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