# exsistance of inverse?

• Oct 3rd 2010, 09:22 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 $\displaystyle f: A \rightarrow B$and $\displaystyle g: B \rightarrow A$ and $\displaystyle g \circ f = 1_A, f \circ g = 1_B$

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

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

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

• Oct 3rd 2010, 11:10 PM
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 $\displaystyle f(x) = c$ for some constant $\displaystyle c \neq 0$ The inverse would be $\displaystyle g= 1/c$
• Oct 4th 2010, 02:38 AM
Defunkt
Quote:

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

No!

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

@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, 03:50 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, 06:16 AM
matt.qmar
aha, ok! so, (not via contradiction though),
for $\displaystyle m,n \in A$
Assume $\displaystyle f(m) = f(n)$
then $\displaystyle g \circ (f(m)) = g \circ (f(n))$
and since $\displaystyle g \circ f = 1_A$
then $\displaystyle 1_A(m) = 1_A(n)$
implies $\displaystyle m=n$
since $\displaystyle m,n \in A$

So f is 1-1

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

(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, 06:54 AM
Defunkt
What properties does an inverse function have?
What properties does g have?
• Oct 4th 2010, 08:42 AM
HallsofIvy
Show that the inverse of f is the unique function such that $\displaystyle f(f^{-1}(x))= x$ and $\displaystyle f^{-1}(f(x))= x$. Since you are told that g has those properities, it [b]is[/b the inverse.
• Oct 4th 2010, 09:25 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 $\displaystyle f: A \rightarrow B$and $\displaystyle g: B \rightarrow A$ and $\displaystyle g \circ f = 1_A, f \circ g = 1_B$

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

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

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

1) From $\displaystyle g \circ f = 1_A$, g is acting as a left inverse of f and f is acting as a right inverse of g.
2) From $\displaystyle f \circ g = 1_B$, f is acting as a left inverse of g and g is acting as a right inverse of f.
So on the one hand, you can say g = $\displaystyle f^{-1}$, the unique two-sided inverse of f.
On the other hand, you can say f = $\displaystyle g^{-1}$, the unique two-sided inverse of g.