Question

(g ◦ f)^-1

g and f are functions. could this be rewritten as g^-1 ◦ f^-1?

Answer 1

h(x) = f o g


h(x) = f(g(x))


y = f(g(x))


x = f(g(y))


f^(-1)(h(x)) = g(y)


g^(-1)(f^(-1)(h(x))) = y


y = g^(-1)(f^(-1)(h(x)))


h^(-1)(x) = g^(-1)(f^(-1)(h(x)))


h^(-1)(x) = g^(-1) o f^(-1)

Answer 2

So if

h(x) = f o g

then

h^(-1)(x) = g^(-1) o f^(-1)

Answer 3

notice the flip

Answer 4

functions, numbers, elements of a group, it makes no difference
\[(ab)^{-1}=b^{-1}a^{-1}\] and the check is that
\[abb^{-1}a^{-1}=aa^{-1}=I\]

Answer 5

Thanks guys I've got this question
Consider the functions given by
f = {(1, 1),(2, 1),(3, 1),(4, 3),(5, 2)} from {1, 2, 3, 4, 5} to {1, 2, 3, 4}
g = {(1, 2),(2, 1),(3, 4),(4, 3)} from {1, 2, 3, 4} to {1, 2, 3, 4, 5}

and I need to find (g ◦ f)^-1 and say if its a function or not so is it correct to write this as g^-1 (f^-1)?

Answer 6

(g ◦ f)^-1 = f^-1 o g^-1

using the idea I showed above

Answer 7

notice this:

say you plug in x = 2 into g^-1

the output will be g^(-1)(2) = 1 using the table above

So f^-1 o g^-1 will turn into f^-1(1)

but

f^-1(1) = {1, 2, 3}

which means that (g ◦ f)^-1 = f^-1 o g^-1 is NOT a function

Answer 8

Thank you so much for your help

Answer 9

are you sure this is correct? why is f^-1(1) = {1, 2, 3} not correct? surely this is like having multiple roots in the equation

Answer 10

notice how when x = 1, f(x) = 1

when x = 2, f(x) = 1

So this suggests that f is not one-to-one

Answer 11

which means that the inverse of f is not a function

Answer 12

so if i plug g^-1 into f^-1 i get = {(1,1), (1,2), (3,3), (1,4)}
this is not a function but how do i phrase it properly because we have to explain why.

Answer 13

if x = 2, then plugging x = 2 into the inverse of g, you get 1

do you see this?

Answer 14

yes I see what you are saying so is writing (g ◦ f)^-1(1) = {1, 2, 4} good enough of an explanation?

Answer 15

yes, you can say that because one input gives you multiple outputs means it's not a function

Answer 16

when you said 'So f^-1 o g^-1 will turn into f^-1(1)'
but in here you could also have 2 or 3 not just 1.

Answer 17

no, you could have 2 or 3 at the end

f^-1(1) is just an intermediate step

Answer 18

the inverse is f^-1 = {(1, 1),(1, 2),(1, 3),(4, 3),(5, 2)}
so if you plug 1 into here you get multiple results

Answer 19

exactly

Answer 20

but my step f^-1(1) is the step right before you get those multiple outputs