$$f^{-1}(S\cup T) = f^{-1} (S)\cup f^{-1}(T)$$
How does one go about showing this equality? What operations can one apply on functions to show this?
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2I am fairly certain that I wrote the proof for this proposition at least three times on the pages of this site. – Asaf Karagila Jan 31 '13 at 23:43
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$x\in f^{-1}(S\cup T)$
$\iff f(x)\in S\cup T$
$\iff f(x)\in S$ or $f(x)\in T$
$\iff x\in f^{-1}(S)$ or $x\in f^{-1}(T)$
$\iff x\in f^{-1}(S)\cup f^{-1}(T)$
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We have a function $f: X\to Y$, and $S, T$ are subsets of $Y$. The inverse image of a subset of $Y$ is the subset of $X$ that contains the elements that $f$ maps to $Y$.
The elements of $f^{-1}(S\cup T)$ are the elements that $f$ maps to either $S$ or $T$, so if $x \in X$ is in $f^{-1}(S)$ then $f(x) \in S\cup T$, and if $x'\in f^{-1}(T)$ then $f(x')$ is in $S\cup T$ as well. We therefore have $$f^{-1}(S)\cup f^{-1}(T) \subseteq f^{-1}(S\cup T)$$ Now, if there is some $x''$ such that it isn't mapped by $f$ to either $S$ or $T$, then it obviously isn't mapped to $S\cup T$, and isn't in $f^{-1}(S\cup T)$, so we can't have proper inclusion and $$f^{-1}(S)\cup f^{-1}(T) = f^{-1}(S\cup T)$$
guest196883
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Ask yourself, "What does $x\in f^{-1}(S\cup T)$ mean? Appeal to the definition and you will see this is an easy proof.
ncmathsadist
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