# exsistance of inverse?

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• Oct 3rd 2010, 10:22 PM
matt.qmar
exsistance of inverse?
hello!

I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!

For functions f,g and sets A, B

Given $f: A \rightarrow B$and $g: B \rightarrow A$ and $g \circ f = 1_A, f \circ g = 1_B$

where $1_A$ is the function $1_A(a) = a, \forall a \in A$

Then $f^{-1}$ exists and $g = f^{-1}$

To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.

Thanks in advance!
• Oct 4th 2010, 12:10 AM
MacstersUndead
Quote:

Originally Posted by matt.qmar
To show the existence of an inverse requires that f is a bijection?

No. f need not have a 1-1 correspondence. take $f(x) = c$ for some constant $c \neq 0$ The inverse would be $g= 1/c$
• Oct 4th 2010, 03:38 AM
Defunkt
Quote:

Originally Posted by MacstersUndead
No. f need not have a 1-1 correspondence. take $f(x) = c$ for some constant $c \neq 0$ The inverse would be $g= 1/c$

No!

If $f(x) \equiv c, \ g(x) \equiv \frac{1}{c}$ for $c \ne 0$ then $f(g(x)) = f \left( \frac{1}{c} \right) = c$ (but it should equal x!) and for the same reason $g(f(x)) = \frac{1}{c}$.

@matt: By definition, a function has an inverse iff it is a bijection. You can try proving this here by contradiction (twice) - first suppose it is not injective and reach a contradiction, and then suppose it is not surjective and again reach a contradiction.
• Oct 4th 2010, 04:50 AM
HallsofIvy
MacstersUndead, when talking about functions, the "inverse" always refers to the inverse function, the function g(x) such that f(g(x))= x and g(f(x))= x, NOT the "reciprocal". A constant function does NOT have an inverse.
• Oct 4th 2010, 07:16 AM
matt.qmar
aha, ok! so, (not via contradiction though),
for $m,n \in A$
Assume $f(m) = f(n)$
then $g \circ (f(m)) = g \circ (f(n))$
and since $g \circ f = 1_A$
then $1_A(m) = 1_A(n)$
implies $m=n$
since $m,n \in A$

So f is 1-1

Now to show it is onto,
for $b \in B$
$f(g(b)) = 1_B(b) = b$
so for each $b \in B$
there is a $g(b) \in A$
so for each $b \in B, \exists g(b) \in A$
such that $f(g(b)) = b$

(each element of B is mapped onto)

so f is onto,

so we know f is a bijection
so it has an inverse!

now... how do we show the inverse is equal to g?

thanks for the help everyone!!
• Oct 4th 2010, 07:54 AM
Defunkt
What properties does an inverse function have?
What properties does g have?
• Oct 4th 2010, 09:42 AM
HallsofIvy
Show that the inverse of f is the unique function such that $f(f^{-1}(x))= x$ and $f^{-1}(f(x))= x$. Since you are told that g has those properities, it [b]is[/b the inverse.
• Oct 4th 2010, 10:25 PM
PiperAlpha167
Quote:

Originally Posted by matt.qmar
hello!

I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!

For functions f,g and sets A, B

Given $f: A \rightarrow B$and $g: B \rightarrow A$ and $g \circ f = 1_A, f \circ g = 1_B$

where $1_A$ is the function $1_A(a) = a, \forall a \in A$

Then $f^{-1}$ exists and $g = f^{-1}$

To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.

Thanks in advance!

Applying what you've been given to the definition of a one-sided inverse, you can conclude the following:

1) From $g \circ f = 1_A$, g is acting as a left inverse of f and f is acting as a right inverse of g.
2) From $f \circ g = 1_B$, f is acting as a left inverse of g and g is acting as a right inverse of f.

Only a bijection, say f, has a unique two-sided inverse, i.e., a unique function that's both the left inverse and the right inverse of f.

But from 1) and 2), you have that g is both a left and right inverse of f, and that f is both a left and right inverse of g.

So on the one hand, you can say g = $f^{-1}$, the unique two-sided inverse of f.
On the other hand, you can say f = $g^{-1}$, the unique two-sided inverse of g.