Functions:Surjective - Revision history

Khanh: /* Surjections as right invertible functions */

2021-11-07T22:54:38Z

Surjections as right invertible functions

Khanh at 22:53, 7 November 2021

2021-11-07T22:53:52Z

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Khanh at 22:31, 7 November 2021

2021-11-07T22:31:57Z

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Lila at 20:40, 27 September 2021

2021-09-27T20:40:13Z

Lila: Created page with "A function is surjective, or "onto", if for all $b\in B$ , there exists at least one $a\in A$ such that $f(a) = b$ ; that is, every elemen..."

2021-09-27T20:37:25Z

Created page with "A function is surjective, or "onto", if for all <math> b\in B </math>, there exists at least one <math> a\in A </math> such that <math> f(a) = b </math>; that is, every elemen..."

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A function is surjective, or "onto", if for all <math> b\in B </math>, there exists at least one <math> a\in A </math> such that <math> f(a) = b </math>; that is, every element in the codomain is mapped to by at least one element in the domain.

Examples:
* Let <math> A = \{-2, -1, 0, 1, 2\} </math> and <math> B = \{1, 3, 9\} </math>, and let <math>f:A\to B</math> such that <math> f(-2) = 9 </math>, <math> f(-1) = 3 </math>, <math> f(0) = 1 </math>, <math> f(1) = 3 </math>, and <math> f(2) = 9 </math>. Each element of <math> B </math> is mapped to by at least one element of <math> A </math>, so <math> f </math> is surjective.

* For the same <math> A </math> and <math> B </math> as in the previous example, let <math> g:A\to B </math> be a function such that <math> g(-2) = 1 </math>, <math> g(-1) = 1 </math>, <math> g(0) = 1 </math>, <math> g(1) = 1 </math>, and <math> g(2) = 1 </math>. This is not a surjective function since there are elements in the codomain that are not mapped to by any elements of the domain.

* Let <math> f:\R\to\R, f(x) = x^3 </math>. For every <math> y\in\R </math>, there exists <math> x = \sqrt[3]{y} \in\R </math> such that <math> f(x) = (\sqrt[3]{y})^3 = y </math>. So, <math> f </math> is a surjective function.

* Let <math> g:\R\to\R, g(x) = \frac{1}{x} </math>. <math> 0 </math> is in the codomain, but there is no <math> x\in\R </math> such that <math> \frac{1}{x} = 0 </math>. Thus, <math> g </math> is not surjective.

==Resources==
* [https://link-springer-com.libweb.lib.utsa.edu/content/pdf/10.1007%2F978-1-4419-7127-2.pdf Course Textbook], pages 154-164

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 Every function with a right inverse is necessarily a surjection. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice.
-If ''f'' : ''X'' → ''Y'' is surjective and ''B'' is a subset of ''Y'', then ''f''(''f''<sup> −1</sup>(''B'')) = ''B''. Thus, ''B'' can be recovered from its preimage {{Nowrap|''f''<sup> −1</sup>(''B'')}}.
+If ''f'' : ''X'' → ''Y'' is surjective and ''B'' is a subset of ''Y'', then ''f''(''f''<sup> −1</sup>(''B'')) = ''B''. Thus, ''B'' can be recovered from its preimage ''f''<sup> −1</sup>(''B'').
 For example, in the first illustration above, there is some function ''g'' such that ''g''(''C'') = 4. There is also some function ''f'' such that ''f''(4) = ''C''.  It doesn't matter that ''g''(''C'') can also equal 3; it only matters that ''f'' "reverses" ''g''.