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Is it possible to have a function which has both left and right inverse but they are unequal ?

A left inverse means the function should be one-to-one whereas a right inverse means the function should be onto.

How can both of these conditions be valid simultaneously without being equal ?

An example will be really helpful. Thanks in advance

Koustav Ghosal
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  • possible duplicate of Right and left inverse, and http://math.stackexchange.com/questions/74363/looking-for-insightful-explanation-as-to-why-right-inverse-equals-left-inverse-f pretty much. – rschwieb Aug 28 '13 at 18:46
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    @rschwieb please notice that the questions you posted are about linear algebra and here is the general case – Dominic Michaelis Aug 28 '13 at 19:05
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    Dear @DominicMichaelis : The initial linked one has no relationship to linear algebra beyond the tag. I don't think your objection applies at all to the first one since the content is identical (and so is the answer.) What you say is true of the second one, though. – rschwieb Aug 28 '13 at 19:22

3 Answers3

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Suppose that $f(g(x))=x$ and $h(f(x))=x$, then $h(x)=h(f(g(x))=g(x)$. So they have to be the same.

robjohn
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For functions they are equal whenever they exist. Let $f$ be a function with left inverse $g$ and right inverse $h$, then $$g=g\cdot id=g(fh)=(gf)h=h$$

Dominic Michaelis
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You can't have mismatched left and right inverses.

If $f$ is a left inverse for $g$ and $h$ is a right inverse for $g$ (denote the identity function $\mathrm{id}(x)=x$) we have $f \circ g = \mathrm{id}$ and $g \circ h = \mathrm{id}$ so $f = f \circ \mathrm{id} = f \circ (g \circ h) = (f \circ g) \circ h = \mathrm{id} \circ h = h$. So $f=h$ is a double sided inverse.

This is true whenever you have an associative operation. Only non-associative operations allow one to have mismatched left and right inverses.

By the way...the equivalence of "existence of a right inverse" and "being onto" assumes the axiom of choice (for those who care about such things).

Bill Cook
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