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Let $f:A\rightarrow B$. Prove each of the following. For some, you will need to assume that $f$ is one-one; for others, that it is onto; for some neither.

  1. $f(\cup A)=\cup f(A)$

  2. $f(\cap A)=\cap f(A)$

  3. $f(A-A_0)=B-f(A_0)$

  4. $f^{-1}(\cup B)=\cup f^{-1}(B)$

  5. $f^{-1}(\cap B)=\cap f^{-1}(B)$

  6. $f^{-1}(B-B_0)=A-f^{-1}(B_0)$

My doubt has to do with knowing when to assume that f has to be one-one of onto or neither. What if I don´t assume the correct property? What is lost in that case? Can you just throw a lot of information here on what you have learned so that I´ll learn more about functions and understand the subject to be able to prove and assume whats needed to solve this. And also in the future this will come to me more naturally?

Henno Brandsma
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Algebry
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1 Answers1

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In order for the inverse of a function to exist it needs to be injective (1-1). So all those problems which involve the inverse require that assumption.

Also, if you learn Latex people will better be able to read your questions and provide answers.

AnotherPerson
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  • And for those which refer to $f^{-1}(B)$ f must be surjective (as B must be the range of f for B to be the domain of f inverse). – fleablood Dec 09 '15 at 11:06
  • Of course not: $f^{-1}[A]$ is always defined, whether $f$ is surjective or injective or not! – Henno Brandsma Dec 09 '15 at 19:07
  • Henno you should take a gander at the inverse function theorem. – AnotherPerson Dec 09 '15 at 20:02
  • @SirJective This has nothing to do with the inverse function theorem, but with the definition of $f^{-1}[A]$ as ${x \in X: f(x) \in A }$. This is defined for all functions $f$ and all subsets $A$ of the codomain. – Henno Brandsma Dec 10 '15 at 18:48
  • Can you please elaborate as to what $f^{-1}(A)$ is when $f$ is not 1-1? – AnotherPerson Dec 10 '15 at 18:50
  • $f(x) = x^2$, A = [1, 4], then $f^{-1}(A) = [-2,-1] \cup [1,2]$. I guess is what Henno is saying. I'm not sure if that is the correct definition but it's consistent. In which my claim $f^{-1}(B)$ implies surjection needn't either. – fleablood Dec 10 '15 at 19:10
  • http://mathworld.wolfram.com/Pre-Image.html "One is not to be mislead by the notation into thinking of the preimage as having to do with an inverse of f" so, yes, Henno is correct, it seems. – fleablood Dec 10 '15 at 19:16
  • But #3 must be surjective to be true. – fleablood Dec 10 '15 at 19:17
  • Sorry just my misunderstanding with the notation. – AnotherPerson Dec 10 '15 at 21:12