Question

Let \(f(A,*) \implies (B, \cdot)\) be homomorphic. Prove that if \(f^{-1}\)is a function then it is homomorphic

Answer 1

work directly from the definition
the hard part is probably figuring out what you need to show, showing it is straight forward
do you know what you need to do?

Answer 2

Well i need to show that it is a bijection?

Answer 3

oh no
you only need to show what it asks you to show
that the inverse function is a homomorphism
you do not need to show that it is a bijection
and in fact it need not be because you only know \(f\) is one to one, not necessarily onto

Answer 4

well it must be onto in order for the inverse to be a function

Answer 5

oh no

Answer 6

it just has to be one to one

Answer 7

ohhhhh noooooooooooooooo

Answer 8

idk i read that somewhere in my bookkk guess i am wrong

Answer 9

think of the map that send \(n\mathbb{Z}\to \mathbb{Z}\) via \(a\to a\)

Answer 10

wait if its not onto then only the range of F can be mapped to A

Answer 11

that is one to one, but certainly not onto
you are only being asked to show that
IF \(f\) is a homo and IF \(f\) is one to one (so the inverse exists) THEN \(f^{-1}\) is a homo

Answer 12

righhhttttttt

Answer 13

that is right only the range can be mapped to A

Answer 14

so the domain of \(f^{-1}\) is the range of \(f\), not necessarily all of \(B\)

Answer 15

ohhhhhhh okkkkkkkkkkkkkk

Answer 16

come up with some examples for yourself
here is one, map \(\mathbb{R}\) to \(\mathbb{R}\times \mathbb{R}\) via \(x\to (x,0)\)
this is certainly one to one, and invertible, but it is not onto for sure

Answer 17

yaaa

Answer 18

in any case your job is to show that \(f^{-1}\) is a homo, that is all
you get to assume it exists, so you get to assume that \(f\) is one to one, and also you are given that \(f\) is homo

Answer 19

write down exactly what you have to show, and exactly what you know

Answer 20

right so that means that \( f(x_1*x_2)=f(x_1)*f(x_2) \)

Answer 21

careful

Answer 22

gotta give me a sec to see how i can pull it all together

Answer 23

ya there shld be that bullet

Answer 24

ok ill rewrite it

Answer 25

\(*\) is the operation in A, \(\cdot\) is the operation in B

Answer 26

\( f(x_1*x_2)=f(x_1) \cdot f(x_2) \)

Answer 27

you get to assume that, yes
you also get to assume one more thing about \(f\)

Answer 28

now \(f^{-1}(y_1*y_2)=f^{-1}(y_1) \cdot f^{-1}(y_2) \)

Answer 29

can we assume that?

Answer 30

no

Answer 31

be careful about where things live
\(x\in A,f(x)\in B\)

Answer 32

\(f^{-1}(y_1)=x_1 \)

Answer 33

ok

Answer 34

i assume you mean "let \(f^{-1}(y_1)=x_1\)"

Answer 35

yaaaaaaaaaaaa

Answer 36

Let \( f^{-1}(y_1)=x_1 \) and\( f^{-1}(y_2)=x_2 \)

Answer 37

don't forget we know one more thing about \(f\) namely that IF \(f(x)=f(x_2)\) THEN \(x_1=x_2\) we may need that later

Answer 38

ok good
now what exactly do you want to show about this

Answer 39

hahah ok this may be incorrect
\( f(f^{-1}(y_1)*f(^{-1}(y_2)) = f(f^{-1}(y_1)) \cdot f(f^{-1}(y_2)) \)

Answer 40

is this totally off track?

Answer 41

i think i is over complicated
write down exactly what it means for \(f^{-1}\) to be a homo, being careful with the multiplication symbols
then i think we can show it, once we know exactly what we need to show

Answer 42

in other words, as usual with a proof, write down precisely what is needed to be shown, working directly from the definition
in other words, translate "\(f^{-1}\) is homomorphic" in to math

Answer 43

ok we need to show that \(f^{-1}(y_1 \cdot y_2) = f^{-1}(y_1)*f^{-1}(y_2) \)