Properties of Kernels of Homeomorphisms

[WWGD]

Hi, let $h: A \rightarrow A$be a homomorphism between algebraic structures. Is there a nice result describing the
properties of $Ker h^2$ , where $h^2 = hoh$ (composition) ? Clearly , $ker( h) \subset ker (h^2 )$ , but are there some other results relating the two; maybe relating kerh to ker h^n?

 

[pasmith]

If x \in \ker h^2 then h(x) \in \ker h (if x \in \ker h then trivially h(x) \in \ker h), so that \ker h^2 = h^{-1}(\ker h). By the same logic \ker h^n = h^{-n+1}(\ker h).

 

[HallsofIvy]

Minor correction: you don't mean "between algebraic structures", you mean "from an algebraic structure to itself".

 

[WWGD]

If x \in \ker h^2 then h(x) \in \ker h (if x \in \ker h then trivially h(x) \in \ker h), so that \ker h^2 = h^{-1}(\ker h). By the same logic \ker h^n = h^{-n+1}(\ker h).

But since $Ker(h) \subset Ker(h^2)$, this would imply both kernels are always equal, and they are not always equal.

Minor correction: you don't mean "between algebraic structures", you mean "from an algebraic structure to itself".

You're right, but I also described the map as being from A to A.

 

[pasmith]

But since $Ker(h) \subset Ker(h^2)$, this would imply both kernels are always equal, and they are not always equal.

It implies nothing of the sort.

If x \in \ker h^2 then by definition <br /> x \mapsto h(x) \mapsto 0.<br /> Thus h(x) \in \ker h, ie. x \in h^{-1}(\ker h). Conversely, if x \in h^{-1}(\ker h) then by definition h(x) \in \ker h, so that h^2(x) = 0, ie. x \in \ker h^2.

Consider the linear map h : \mathbb{R}^2 \to \mathbb{R}^2 with h(1,0) = 0 and h(0,1) = (1,0). Then \ker h is the subspace spanned by (1,0) but \ker h^2 = h^{-1}(\ker h) = \mathbb{R}^2: <br /> (x,y) \mapsto (x,0) \mapsto (0,0).<br />

 

[HallsofIvy]

You're right, but I also described the map as being from A to A.

I said it was a minor correction!

 

[WWGD]

It implies nothing of the sort.

If x \in \ker h^2 then by definition <br /> x \mapsto h(x) \mapsto 0.<br /> Thus h(x) \in \ker h, ie. x \in h^{-1}(\ker h). Conversely, if x \in h^{-1}(\ker h) then by definition h(x) \in \ker h, so that h^2(x) = 0, ie. x \in \ker h^2.

Consider the linear map h : \mathbb{R}^2 \to \mathbb{R}^2 with h(1,0) = 0 and h(0,1) = (1,0). Then \ker h is the subspace spanned by (1,0) but \ker h^2 = h^{-1}(\ker h) = \mathbb{R}^2: <br /> (x,y) \mapsto (x,0) \mapsto (0,0).<br />

O.K, sorry I misread that $ker h^2 \subset ker h$ , which would then give a double inclusion implying equality.

I said it was a minor correction!

Sorry, I was being too nit-picky

 

[platetheduke]

You have that $h$ maps $\ker h^2$ into $\ker h$. This is not injective or subjective in general.