Xyx^-1y^-1 a Lie group homomorphism?

[Sajet]

Hi! I was just going through this script on Lie groups: http://www.mit.edu/~ssam/repthy.pdf

At one point the following is said:

(see attachment)

I've spent multiple hours trying to figure out why this is a group homomorphism. Sure, once you know the theorem is correct, this follows. But without knowing this, I can't figure out why it should be.

I want to see that:

$\varphi_y(x_1x_2) = \varphi_y(x_1)\varphi_y(x_2) \Leftrightarrow x_1x_2yx_2^{-1}x_1^{-1}y^{-1} = x_1yx_1^{-1}y^{-1}x_2yx_2^{-1}y^{-1}$


If I write $x_1y = \bar y_1x_1, x_2y = \bar y_2x_2$, then the above equation is equivalent to

$x_1\bar y_2x_1^{-1} = \bar y_1y^{-1}\bar y_2$

but I don't get any further from here.

 

[Sajet]

Anyways, I think whether or not this is a homomorphism is not necessary for this proof. One can just argue that this map is smooth since G is a Lie group, therefore the image under G is connected, therefore it is a one point set and since eye^-1y^-1 = e one gets that this set must be exactly {e}.

 

[DonAntonio]

Anyways, I think whether or not this is a homomorphism is not necessary for this proof. One can just argue that this map is smooth since G is a Lie group, therefore the image under G is connected, therefore it is a one point set and since eye^-1y^-1 = e one gets that this set must be exactly {e}.

I agree with you: I can't either see how that map is a homomorphism in a general case, but it is unnecessary to consider it that way.

DonAntonio