# Properties of Kernels of Homeomorphisms

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1. Oct 1, 2014

### 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?

2. Oct 2, 2014

### 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 $$\ker h^n = h^{-n+1}(\ker h).$$

3. Oct 2, 2014

### HallsofIvy

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

4. Oct 2, 2014

### WWGD

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

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

5. Oct 3, 2014

### pasmith

It implies nothing of the sort.

If $x \in \ker h^2$ then by definition $$x \mapsto h(x) \mapsto 0.$$ 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 $\ker h^2 = h^{-1}(\ker h) = \mathbb{R}^2$: $$(x,y) \mapsto (x,0) \mapsto (0,0).$$

6. Oct 4, 2014

### HallsofIvy

Staff Emeritus
I said it was a minor correction!

7. Oct 4, 2014

### WWGD

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

Sorry, I was being too nit-picky

8. Oct 10, 2014

### platetheduke

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