1. ## 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.

2. 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 $g= 1/c$

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.

4. 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.

5. 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!!

6. What properties does an inverse function have?
What properties does g have?

7. 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.

8. 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.

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.
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.