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Is this function onto? Remark. This function maps ordered pairs to a single real numbers. The image of an ordered pair is the average of the two coordinates of the ordered pair. To decide if this function is onto, we need to determine if every element in the codomain has a preimage in the domain. Solution. Take any real number, \(x \in \mathbb ...
- 8.4: Images and Preimages
If \(f^{-1}:Y\to X\) is a function, then it is sensible to...
- 8.4: Images and Preimages
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) ".
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 ...
Apr 17, 2022 · If \(f^{-1}:Y\to X\) is a function, then it is sensible to write \(f^{-1}(y)\) for \(y\in Y\). 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 \(T\).
Dec 8, 2012 · 8. I would not use the terminology preimage of a function at all. If f: X → Y f: X → Y is a function, and A ⊆ Y A ⊆ Y, the preimage of A A under f f is. f−1[A] ={x ∈ X: f(x) ∈ A}. f − 1 [A] = {x ∈ X: f (x) ∈ A}. That is, in my terminology subsets of the codomain of f f have preimages under f f, and these preimages are ...
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 ...
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When our function satisfies very specific conditions, we can ensure that the preimage of a set containing a single element is always set containing a single element. Understanding those conditions is one of the main aims of this chapter and we’ll discuss it in detail in the next section. That, in turn, will help us to define the inverse function.
