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)

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Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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