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My textbook asks me to decide whether or not this expression is true:

Given the function $f: X \to Y$ with $B_1 \subseteq Y $. $ f^{-1}(Y $ \ $ B_1) = X $ \ $f^{-1}(B_1) $

I was confused because there is no assumption that the function $f$ is onto. Hence there might be an element $y \in Y$ such that for no $x \in X$, $f(x) = y$. I actually don't know if in this case the expression $f^{-1}(Y)$ makes sense.

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    $\begingroup$ By definition, $f^{-1}(Y)=\{x\in X\mid f(x)\in Y\}=X$. $\endgroup$ – Ittay Weiss Jul 1 '14 at 10:14
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One Interpretation (probably wrong):

If $f: X \rightarrow Y$ and you assume that $f^{-1}$ exists as a mapping at element level from $Y \rightarrow X$, then $f$ must be not only surjective but also injective. From that you can then show $f^{-1}(Y \setminus B_1) = X \setminus f^{-1}(B_1)$.

Another Interpretation (probably correct):

Using the suggested notataion in comments, $f^{-1} [B_1] = \{x \in X\colon f(x) \in B_1\}$ and $f^{-1} [Y] = \{x \in X\colon f(x) \in Y\}= X$ (assuming that $f$ is follows the normal definition of a fucntion from $X$).

$$\begin{align} f^{-1} [Y \setminus B_1] &= \{ x \in X\colon f(x) \in Y \setminus B_1\}\\ &=\{ x \in X\colon f(x) \in Y \land f(x) \notin B_1\}\\ &= \{ x \in X\colon x \in X \land x \notin f^{-1} [B_1]\}\\ &= X \setminus f^{-1} [B_1]. \end{align}$$

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  • $\begingroup$ You're misinterpreting the symbol $f^{-1}(B)$ where $B\subseteq Y$. Admittedly, the notation is ambiguous. By definition $f^{-1}(B)=\{x\in X\mid f(x)\in B\}$. You shouldn't look at $f^{-1}(B)$ as the preimage of the inverse of $f$, but rather as string of symbols which happens to have the symbol $f^{-1}$ in it. Personally I prefer the notation $f^{-1}[B]$. $\endgroup$ – Git Gud Jul 1 '14 at 11:46
  • $\begingroup$ @Git Gud - I learn't something there. $\endgroup$ – Tom Collinge Jul 1 '14 at 12:42
  • $\begingroup$ @GitGud My preference goes to $f^\gets$ (inverse image) and $f^\to$ (direct image). $\endgroup$ – egreg Jul 1 '14 at 13:47
  • $\begingroup$ @egreg It's a nice notation. I've never seen it before. $\endgroup$ – Git Gud Jul 1 '14 at 13:49
  • $\begingroup$ @GitGud Probably somebody would use $f_\to$ (it's applying a covariant functor). A teacher of mine used these arrows, I don't remember who. In any case, I find $f^{-1}(B)$ very confusing for the students. Using $[]$ around the set helps, though, provided one has talked about relations and sections. $\endgroup$ – egreg Jul 1 '14 at 13:59

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