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ONT Re: Logic Of Relatives




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LOR.  Note 52

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In the case of a 2-adic relation F c X x Y that has
the qualifications of a function f : X -> Y, there
are a number of further differentia that arise:

| f is "surjective"   iff   f is total at Y.
|
| f is "injective"    iff   f is tubular at Y.
|
| f is "bijective"    iff   f is 1-regular at Y.

For example, or more precisely, counterexample,
the function f : X -> Y that is depicted below
is neither total at Y nor tubular at Y, and so
it cannot enjoy any of the properties of being
sur-, or in-, or bi-jective.

0   1   2   3   4   5   6   7   8   9
o   o   o   o   o   o   o   o   o   o   X
|    \  |  /     \   \  |   |    \ /
|     \ | /       \   \ |   |     \     f
|      \|/         \   \|   |    / \
o   o   o   o   o   o   o   o   o   o   Y
0   1   2   3   4   5   6   7   8   9

A cheap way of getting a surjective function out of any function
is to reset its codomain to its range.  For example, the range
of the function f above is Y'=  {0, 2, 5, 6, 7, 8, 9}.  Thus,
if we form a new function g : X -> Y' that looks just like
f on the domain X but is assigned the codomain Y', then
g is surjective, and is described as mapping "onto" Y'.

0   1   2   3   4   5   6   7   8   9
o   o   o   o   o   o   o   o   o   o   X
|    \  |  /     \   \  |   |    \ /
|     \ | /       \   \ |   |     \     g
|      \|/         \   \|   |    / \
o       o           o   o   o   o   o   Y'
0       2           5   6   7   8   9

The function h : Y' -> Y is injective.

0       2           5   6   7   8   9
o       o           o   o   o   o   o   Y'
|       |            \ /    |    \ /
|       |             \     |     \     h
|       |            / \    |    / \
o   o   o   o   o   o   o   o   o   o   Y
0   1   2   3   4   5   6   7   8   9

The function m : X -> Y is bijective.

0   1   2   3   4   5   6   7   8   9
o   o   o   o   o   o   o   o   o   o   X
|   |   |    \ /     \ /    |    \ /
|   |   |     \       \     |     \     m
|   |   |    / \     / \    |    / \
o   o   o   o   o   o   o   o   o   o   Y
0   1   2   3   4   5   6   7   8   9

Jon Awbrey

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