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Inverse functions and null set.

by blastoise
Tags: functions, inverse, null
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blastoise
#1
Apr18-12, 07:10 PM
P: 22
Ok,

I understand an inverse function sends a variable in the range to the corresponding value in the domain, but am not sure if what I'm thinking is correct... : For example:

Let A be the set


[itex] A = \{1,2,3,7,8\} ; B = \{4,5,6\} [/itex] and the function [itex] f [/itex] map A to B s.t

f(1) = 4
f(2) = 5
f(3) = 6

so 7,8 do not have a value that is mapped one to one.

I understand f is an surjection, but not a injection . But, does an inverse function exist?

I would say yes, despite there not being a value in B that maps to 7 or 8.
Is my thinking correct?


Also, am I correct to say that a function does not have to use every element in the domain in order to have an inverse; I am confused because wouldn't one just say it maps to the null set and the inverse of the null set would contain values(7,8) that it maps to and hence not a function...?
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DonAntonio
#2
Apr18-12, 09:08 PM
P: 606
Quote Quote by blastoise View Post
Ok,

I understand an inverse function sends a variable in the range to the corresponding value in the domain, but am not sure if what I'm thinking is correct... : For example:

Let A be the set


[itex] A = \{1,2,3,7,8\} ; B = \{4,5,6\} [/itex] and the function [itex] f [/itex] map A to B s.t

f(1) = 4
f(2) = 5
f(3) = 6

so 7,8 do not have a value that is mapped one to one.


*** ...and thus f is NOT a function from A to B but one from a proper subset of A to B...***



I understand f is an surjection, but not a injection . But, does an inverse function exist?


*** Yes, if we look at f defined from {1,2,3} onto {4,5,6}. ***



I would say yes, despite there not being a value in B that maps to 7 or 8.
Is my thinking correct?


Also, am I correct to say that a function does not have to use every element in the domain in order to have an inverse; I am confused because wouldn't one just say it maps to the null set and the inverse of the null set would contain values(7,8) that it maps to and hence not a function...?

If we talk of a function [itex]f: A\to B[/itex] , we're explicitly assuming f is defined in the whole of A. Whether f is 1-1 and/or onto is another matter.

DonAntonio


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