Fruits of procrastination

-September 1, 2013September 1, 2013Uncategorized

Let $f:X\to Y$ be a mapping. We will prove that $f^{-1}(Y-f(X-W))\subseteq W$ , with equality when $f$ is injective. Note that $f$ does not have to be closed, open, or even continuous for this to be true. It can be any mapping.

Let $W\subseteq X$ . The mapping of $W$ in $Y$ is $f(W)$ . As for $f(X-W)$ , it may overlap with $f(W)$ , we the mapping be not be injective. Hence, $Y-f(X-W)\subseteq f(W)$ .

>Taking $f^{-1}$ on both sides, we get $f^{-1}(Y-f(X-W))\subseteq W$ .

How can we take the inverse on both sides and determine this fact? Is the reasoning valid? Yes. All the points in $X$ that map to $Y-f(X-W)$ also map to $W$ . However, there may be some points in $f^{-1}(W)$ that do not map to $Y-f(X-W)$ .

Are there other analogous points about mappings in general? In $Y$ , select two sets $A$ and $B$ such that $A\subseteq B$ . Then $f^{-1}(A)\subseteq f^{-1}(B)$