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Oct 10, 2024 · Let f:A->B be a map between sets A and B. Let Y subset= B. Then the preimage of Y under f is denoted by f^(-1)(Y), and is the set of all elements of A that map to elements in Y under f. Thus f^(-1)(Y)={a in A|f(a) in Y}. (1) One is not to be mislead by the notation into thinking of the preimage as having to do with an inverse of f. The preimage is defined whether f has an inverse or not. Note ...
- Preimage
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- Preimage
In mathematics, for a function , the image of an input value is the single output value produced by when passed . The preimage of an output value is the set of input values that produce . More generally, evaluating at each element of a given subset of its domain produces a set, called the " image of under (or through) ".
So, given an arbitrary element of the codomain, we have shown a preimage in the domain. Thus every element in the codomain has a preimage in the domain. Therefore, by the definition of onto, \(g\) is onto.
Mathematics help chat. Mathematics Meta ... As a consequence of this we get that the preimage of complement is complement of the preimage, see here: ...
The preimage of an element y in set B is the set of all elements x in set A such that f(x) = y. The preimage is denoted as f^(-1)(y), where the (-1) superscript does not denote the inverse function but represents the preimage operation. To visualize this, let’s consider a simple function f(x) = x^2, which maps elements from set A (real ...
Oct 10, 2024 · Wolfram for Education. Created, developed and nurtured by Eric Weisstein at Wolfram Research. Given f:X->Y, the image of x is f (x). The preimage of y is then f^ (-1) (y)= {x|f (x)=y}, or all x whose image is y. Images are elements of the range, while preimages are subsets (possibly empty) of the domain.
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Apr 17, 2022 · Notice that we defined the preimage of a subset of the codomain regardless of whether \(f^{-1}\) is a function or not. In particular, for \(T\subseteq Y\) , \(f^{-1}(T)\) is the set of elements in the domain that map to elements in \(T\) .
