9
$\begingroup$

Suppose that $f:X \to Y$ is an onto function. Prove that for all subsets $B$ subset of $Y$, $f(f^{-1}(B)) = B$. I don't know how to do this if the function is not also one to one, which it is not. Any help proving this would be greatly appreciated.

$\endgroup$
5
$\begingroup$
  • $\exists$ is the symbol for "there exists."
  • $\wedge$ is the symbol for "and."
  • $f:X\twoheadrightarrow Y$ is notation for "$f$ is a surjective (onto) function from $X$ to $Y$."
  • $\mathcal{P}(Y)$ is notation for "powerset of $Y$" or "the set of all subsets of $Y$."

  • "$\subseteq$" becomes "$=$" for onto functions.
| cite | improve this answer | |
$\endgroup$
  • 5
    $\begingroup$ which book did you get the picture from? $\endgroup$ – ak87 Feb 21 '17 at 18:11
4
$\begingroup$

Note that:

  1. For any $y\in f(f^{-1}(B)$, there exists $x \in f^{-1}(B)$ satisfying $y=f(x)$. Clearly $y=f(x) \in B$.

  2. For any $y\in B$, since $f(x)$ is an onto function, there is $x\in X$ satisfying that $f(x)=y$, i.e.,$x\in f^{-1}(y)$. So $y=f(x)\subset f(f^{-1}(y)) \subset f(f^{-1}(B)).$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I would further elaborate on second point why it is true. If f is not an onto function , then it does not necessary mean there exists x satisfying $f(x)=y$. For this case B will never be a subset of $f(f^-1(B))$ , that’s why the fact that f is onto function is important. $\endgroup$ – Ling Min Hao Jul 24 '18 at 20:02
1
$\begingroup$

Given $f : X \to Y$, and $\forall y \in Y \exists x \in X (f(x) = y)$.

Want to show that $\forall B \subset Y, f(f^{-1}(B)) = B$

We know that $f^{-1}(B) = \{ b : f(b) \in B \}$. If this set is never empty, then we have our result. The set is never empty by our second assumption.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

First, $f^{-1}(y) = \lbrace x \in X : f(x) = y)\rbrace $. If $y \in B$, so exist $x \in X$ that $f(x) = y$ (f is onto), then $f(f^{-1}(y)) \in B$, by definition, so $f(f^{-1}(y)) \subset B$. Now, $y \in f(f^{-1}(B))$, how f is onto exist $x \in f^{-1}(B)$, such that $f(x) = y$, then $y \in B$, so $f(f^{-1}(B)) \subset B$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.