For discussion.

Let A and B be two sets and f a function from A to B. If C is a subset of the range B then the preimage, or inverse image, of C under the function f is the set defined as f

^{-1}(C) = {x belong to A : f(x) belong to C }

http://www.mathcs.org/analysis/reals/logic/relation.htmlI think C is not necessary a subset of the range B. C can be any set as long as" f

^{-1}(C) = {x belong to A : f(x) belong to C } "

Otherwise, we can not solve this question by "

http://www.mathcs.org/analysis/reals/cont/answers/topcont2.html"

Let f(x) = 1 if x > 0 and f(x) = -1 if x 0. Show that f is not continuous by

1.finding an open set whose inverse image is not open.

2.finding a closed set whose inverse image is not closed.

the range of f(x) is 1 and -1, and are only these two numbers, but (-2,0) is not a subset of B. If we use above inverse image definition even the question "find an open set" itself is not correct.

So I think the domain of inverse image is not necessarily a subset of rang f.