# The preimage of a subset

If $A\subseteq B$ under what conditions is $f^{-1}(A) \subseteq f^{-1}(B)$, where $f^{-1}$ is the preimage, not the inverse.

First, we should be clear about the definition of $f$. Suppose $X, Y$ are sets and $f: X \to Y$ is a map. Suppose $A, B \subseteq Y$ with $A \subseteq B$.

It's always true that $f^{-1}(A)$ is contained in $f^{-1}(B)$, and this should be clear by the definition of preimage.

$f^{-1}(A)$ is the stuff in $X$ that is mapped into $A$. But since $A \subseteq B$, if stuff in $X$ is mapped into $A$, then that same stuff is mapped into $B$ because $A$ is a subset of $B$. Then that means the stuff mapped into $A$ is a subset of the stuff mapped into $B$, i.e., $f^{-1}(A) \subseteq f^{-1}(B)$.

Bonus question for you: If $A \subseteq B$, when is $f^{-1}(A) = f^{-1}(B)$?

• If and only if $A=B$? – usainlightning Feb 10 '15 at 19:05
• $f^{-1}(A) =f^{-1}(B) \iff${$x \in \mathbb{R}|f(x) \in A \cap B$}, Therefore as long as $f^{-1}(A/B)$ and $f^{-1}(B/A)$ is the empty set this is true. – usainlightning Feb 10 '15 at 19:51
• I didn't see your hint, am i still correct? – usainlightning Feb 10 '15 at 19:52
• @usainlightning I'm not sure what you mean with your $\iff$. You have a statement on the left hand side, but a set on the right hand side. Here is what I was looking for as a proof: By my hint, since $B = A \cup (B - A)$, then $f^{-1}(B) = f^{-1}(A) \cup f^{-1}(B - A)$. So if $f^{-1}(B - A) = \emptyset$, then we get $f^{-1}(B) = f^{-1}(A) \cup \emptyset = f^{-1}(A)$. – layman Feb 10 '15 at 20:32
• Yes sorry that is obvious – usainlightning Feb 10 '15 at 21:14