$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$

Question

When $f$ and $g$ are invertible functions we have $$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$$

I think we have this in every Group $(A,*,u,')$

$(a*b)=c$

$(a*b)*b'=a=c*b'$

$c'*a=c'*(c*b')=b'$

$(c'*a)*a'=b'*a'$

$$c'=(a*b)'=b'*a'$$

and in general we have

$(a_1*a_2*...*a_n)'=a_n'*...*a_2'*a_1'$

1)What is the specific name of this theorem?

2)There is something wrong with my derivation?

EDIT Can we see this as a property of a function $f:A\rightarrow A$ on a structure $(A,\circ)$:

$f(x\circ y)=f(y)\circ f(x)$

?

This is the inverse property in group theory. – Easy Apr 20 '13 at 16:59 — Easy, Apr 20 '13 at 16:59
You could say $a\mapsto a'$ is an antiisomorphism. – xavierm02 Apr 20 '13 at 17:01 — xavierm02, Apr 20 '13 at 17:01
@xavierm02 thanks that what I was looking for – m_l Apr 20 '13 at 17:39 — m_l, Apr 20 '13 at 17:39

score 1 · Answer 1 · answered Apr 20 '13 at 17:02

1

This is not really a theorem, just a trivial fact that's easily obtained from the definition of an inverse and associativity.

$$(f \circ g) \circ (g^{-1} \circ f^{-1}) = f \circ (g \circ g^{-1}) \circ f^{-1} = f \circ f^{-1} = 1$$

answered Apr 20 '13 at 17:02

Alex Provost

yea, that derivation is very easier. now seems moore trivial. – m_l Apr 20 '13 at 17:33

score 1 · Accepted Answer · answered Apr 20 '13 at 17:04

1) There isn't one as far as I know! It's just a well known fact. See e.g. http://www.proofwiki.org/wiki/Inverse_of_Matrix_Product

2) Your derivation is a little roundabout, but perfectly correct. I'd just write $ab\cdots c c'\cdots b'a' = \iota$.

2 Answers2