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I was solving a problem of set theory and applications then suddenly I faced this problem:

Let $ f: X \to Y $ be a function, and let $ B \subseteq Y $. Prove that: \begin{align*} f^{-1}(Y-B) = X - f^{-1}(B) \end{align*}

I proved that:

\begin{align*} f^{-1}(Y-B) = f^{-1}(Y) - f^{-1}(B) \end{align*}

Now I think It is enough to show that $ f^{-1}(Y) = X $.
A friend of mine told me that the equality $ f^{-1}(Y) = X $ is always true, but I have doubt. What does that mean? Is it really true that the pre-image of a co-domain is always the domain?! if not what is the counterexample?

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  • $\begingroup$ For $f$ to be a function and $X$ to be its domain, every point of the domain must be mapped to a point in the co-domain. Therefore if you try to compute the "inverse" of any point in the co-domain, it will either not exist and if it does, it has to be in the domain. Since all points in the domain have an image in the co-domain, you get what you want. $\endgroup$ – Iguana Oct 28 '19 at 7:19
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Remember what it means for $f$ to be a function from $X$ to $Y$. Since $X$ is the domain of $f$ and $Y$ is the codomain, for every value $x\in X$, we have that $f(x)=y$ for some $y\in Y$. This is just what it means for $f$ to be a function from $X$ to $Y$, or for $f$ to "pass the vertical line test" as it is sometimes said in introductory courses.

When we say $f^{-1}(Y)$, we're talking about the subset of elements in the domain $X$ that get mapped into $Y$ by $f$. But every element in $X$ gets mapped into $Y$ by $f$ because that's precisely what it means for $f$ to be a function from $X$ to $Y$. Your friend is certainly correct that $f^{-1}(Y)=X$.

In fact, the image of $f$, $\mathrm{Im}(f)\subset Y$ has the property that $f^{-1}(\mathrm{Im}(f)) = X$, so any set that contains the image also has this property (this includes $Y$). If it helps, maybe you can think of this as being because $f^{-1}(Y) = f^{-1}\left(\mathrm{Im}(f) \cup (Y-\mathrm{Im}(f))\right) = f^{-1}(\mathrm{Im}(Y)) \cup f^{-1}(Y-\mathrm{Im}(Y)) = X \cup \emptyset = X$.

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    $\begingroup$ Thanks to you, I got it, but I didn't understand the last line you wrote about the image of $Y$, and if I suppose you meant to write the pre-image of $Im(f)$ why it would be $ Y $ ? I think you should've written $X$ instead. $\endgroup$ – Reza Oct 28 '19 at 7:47
  • $\begingroup$ @Reza Oh, yes, this was a typo! My bad. You are correct. $\endgroup$ – Jack Crawford Oct 28 '19 at 7:48
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Let $x\in f^{-1}(Y-B)$ then $\exists$some $y\in Y-B$ such that $y=f(x)$ Now $f(x)\notin B$ so $x\notin f^{-1}(B)$ so $x \in X-f^{-1}(B)$.Now conversely$x\notin f^{-1}(B)$ implies $f(x)\notin B$ so $f(x)\in Y-B$ thus $x\in f^{-1}(Y-B)$.Trivial case where $f^{-1}(Y-B)$ is empty is not discussed.I think you can do it from above. And for your doubt,check the definition, $f^{-1}(Y)=\{x\in X : f(x) \in Y\}$ which is obviuously $X$,because for any element of $X$,the property is true.

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  • $\begingroup$ Thanks for your solution ! $\endgroup$ – Reza Oct 28 '19 at 7:50

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