Prove $f^{−1}(T − Y ) = S − f^{-1}(Y )$.

Question

I could use some help here:

Let $f:S\rightarrow T$ be a function. If $Y \subseteq T$, we define the inverse image of $Y$ to be $f^{−1} (Y ) = \{x ∈ S : f(x) ∈ Y \}$. Prove $f^{−1}(T − Y ) = S − f^{-1}(Y )$.

I start by supposing $x\in f^{-1}(T-Y)\subseteq S$ which gives $f(x)\in T-Y$, and it seems like $f(x)$, the set $T$, and the set $S$ can somehow be transformed into $x$, $S$, and $f^{-1}(Y)$ respectively via $f^{-1}$, but I'm not sure how to do this. Doing so would prove $f^{−1}(T − Y ) \subseteq S − f^{-1}(Y )$ and then doing the proof in reverse would show the reverse, thus proving equality.

Answer 1

Let $y\in f^{-1}(T-Y)$. Since $$\begin{array} 1f^{-1}(T-Y)&=\{s\in S: f(s)\in T-Y\}\\&=\{s\in S:f(s)\in T\text{ and }f(s)\notin Y\}\\&=f^{-1}(T)-f^{-1}(Y)\\&=S-f^{-1}(Y)\end{array}$$

Answer 2

The following statements are equivalent for a function $f:S\to T$:

• $x\in f^{-1}(T-Y)$
• $f(x)\in T-Y$
• $f(x)\notin Y$
• $x\notin f^{-1}(Y)$
• $x\in S-f^{-1}(Y)$

Looking at the first and last bullet we can conclude that:$$f^{-1}(T-Y)=S-f^{-1}(Y)$$

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To be well equipped for problems like this put the following rule in your math luggage:$$x\in f^{-1}(A)\iff f(x)\in A$$