Question

f: A -> B and g: B -> C are 1-1 functions. Prove that the inverse of g composition f = f inverse composition g inverse on Ran(g composition f)

Answer 1

\[f: A \rightarrow B \] and \[g: B \rightarrow C \]
are one-to-one.
Prove that \[[g(f(x))]^{-1} = f^{-1}(g^{-1}(x))\] on
\[Ran(g(f(x))\]
(someone has to show me how to put all of these on the same line and how to use the little circle for composition D: )

Okay, so here's my brainstorming. I at least want to say what I'm thinking and what I've tried.

Okay, so if I say:
\[f^{-1}= \left\{ (y,x) \in B \times A | f(x) = y \right\}\]
and then I say:
\[(g(f(x)) = \left\{ (x,z) | z = g(y) , y = f(x), x \in A\right\}\] Does this mean I can say:
\[(g(f(x))^{-1} = \left\{ (z,x) | x = g(y), y= f(z), z \in C \right\}\]???

Assuming that's correct, in order to show equality, I need to show:
\[(g(f(x))^{-1} \subseteq f^{-1}(g^{-1}(x))\] AND
\[f^{-1}(g^{-1}(x)) \subseteq g(f(x))^{-1}\] correct?

One thing I'm definitely not sure on is when it says on Ran(g(f(x)). I guess something like this only holds when we consider the set Ran(g(f(x))?

Alright, so I'm assuming I have to say something like let
\[z \in (g(f(x))^{-1}\] What that implies, Im not sure. That there exists x = g(f(z)), z in C?

I know I'm leaving out the idea of one-to-one, but I just don't know how to have one thing imply another or how to connect the ideas together. All I have written down are a bunch of set definitions with no idea how to piece them together :(

Answer 2

\circ

Answer 3

it is always the case that the inverse of \(ab\) is \(b^{-1}a^{-1}\) not matter what the group is

Answer 4

the only reason we don't notice it is because multiplication is commutative, so it doesn't matter what order you write it, but
\[(ab)(b^{-1}a^{-a})=a(bb^{-1})a^{-1}\] as composition is associative, then
\[aIa^{-1}\] as \(bb^{-1}=I\) then
\[aa^{-1}=I\] and done

Answer 5

to repeat, has nothing to do with composition as such, true in any associative group

Answer 6

So, multiplication as an operator in any group is commutative and composition as an operator in any group is associative?

And do I not have to show that \[(ab)^{-1} = b^{-1}a^{-1}\] I can just state that?

Answer 7

@satellite73

Answer 8

no i did not say it was commutative, and in fact it is not

Answer 9

associativity is what is needed to write
\[(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}\] that is all

Answer 10

Oh, you used the word commutative above, maybe you just meant associative then, I'm sorry. So do I need to mention anything about one-to-one or about the Ran(g(f(x)) in any of this?

Answer 11

in fact if your operation is associative, then the parentheses are not needed, and the one line proof that \((ab)^{-1}=b^{-1}a^{-1}\) is
\[abb^{-1}a^{-1}=aea^{-1}=aa^{-1}=e\]

Answer 12

when i wrote
"the only reason we don't notice it is because multiplication is commutative"
i meant with numbers for example it doesn't matter
\[(ab)^{-1}=a^{-1}b^{-1}\] since multiplication is commutative we can write the inverse in any order we like

Answer 13

but to show that that works at one point you have to use commutativity
not so with \((ab)^{-1}=b^{-1}a^{-1}\) it just works out of the box

Answer 14

I see. Now I just want to make sure I dont need to make mention of 1-1 or the range portion of the problem.

Answer 15

maybe you do
i think that just makes the composition make sense is all

Answer 16

Alright. Ill see if I can incorporate it in there. Thanks for you help ^_^

Answer 17

yw