Question

Prove that these functions are NOT an isomorphism:
1/ f : Z→Z6 defined by f(x) = x mod 6
2/ f : {0}→Z defined by f(x) = x
3/ A : R2 →R2 defined by A =[1 0
0 2]

Answer 1

if a function is an isomorphism then it is:
1/ one to one
2/ onto
3/ f(a+b)=f(a)+f(b), f(a.b)=f(a).f(b)

Answer 2

so one of these properties must fail!

Answer 3

For number 1,
Determine f(1) and also f(7).
Do you see any problems?

Answer 4

7 does not belong to Z6

Answer 5

Well we're taking the 7 from Z, our domain.
So it's ok to plug 7 into the function.
But what are we getting OUT when plug 7 IN?

f(7)=7mod6=1
yes? :)

Answer 6

But we also have
f(1)=1mod6=1

This breaks one of our rules since we clearly can see that f(1)=f(7).
Which one? :)

Answer 7

the first one! since 1 is not equal 7, right?

Answer 8

Yess good good good.
We don't have one-to-one'ness

Answer 9

yeah

Answer 10

Any ideas what's wrong with the next one?