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Topology questions - sets and functions

Topology
Sets and Functions (XXXV)
Functions

Let X be a non-empty set. The identity mapping ix on X is the mapping of X onto itself
defined by ix(x) = x for every x. Thus ix sends each element of X to itself; that is,
it leaves fixed each element of X. If f is one-to-one onto, so that its inverse f- 1 exists,
show that f f- 1 = f- 1 f = ix .

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This solution is comprised of a detailed explanation to show that f f- 1 = f- 1 f = ix .
It contains step-by-step explanation of the following problem:

Let X be a non-empty set. The identity mapping ix on X is the mapping of X onto itself
defined by ix(x) = x for every x. Thus ix sends each element of X to itself; that is,
it leaves fixed each element of X. If f is one-to-one onto, so that its inverse f- 1 exists,
show that f f- 1 = f- 1 f = ix .

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