I'm reading through James Munkres' book, Topology, and the theorem about the image of a connected space under mapping being connected.
To paraphrase the proof:
Let $f:X\to f(X)$ be a continuous map; let $X$ be connected.
Suppose that $f(X)=A\bigcup B$ is a separation of $f(X)$ into two disjoint nonempty sets open in $f(X)$. Then $f^{-1}(A)$ and $f^{-1}(B)$ are disjoint sets whose union is $X$. (The proof continues but I'll stop here.)
My question is why are $f^{-1}(A)$ and $f^{-1}(B)$ disjoint in $X$? I've seen that $f^{-1}(A\cap B)\subset f^{-1}(A)\cap f^{-1}(B)$, but I don't think this helps as I want to know $f^{-1}(A)\cap f^{-1}(B)=\emptyset$ from the fact that $A\cap B=\emptyset$.
$\endgroup$2 Answers
$\begingroup$If $x\in f^{-1}(A)$ then $f(x)\in A$ and since $A$ and $B$ are disjoint $f(x)\not\in B$. Therefore no element of $A$ can be in $B$ and the same argument shows that no element of $B$ can be in $A$ so they are disjoint.
$\endgroup$ $\begingroup$You don't actually need any special conditions or much reasoning at all here; this is a consequence of the requirement that a function match each input to a single output.
Imagine the claim were false. Then there must be some $x$ in the preimages of both $A$ and $B$. This is impossible, because $x$ can only be mapped to a single value, but $A$ and $B$ are disjoint.
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