Properties of Kernels of Homeomorphisms

  • #1
WWGD
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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?
 

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  • #2
pasmith
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If [itex]x \in \ker h^2[/itex] then [itex]h(x) \in \ker h[/itex] (if [itex]x \in \ker h[/itex] then trivially [itex]h(x) \in \ker h[/itex]), so that [tex]\ker h^2 = h^{-1}(\ker h).[/tex] By the same logic [tex]\ker h^n = h^{-n+1}(\ker h).[/tex]
 
  • #3
HallsofIvy
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Minor correction: you don't mean "between algebraic structures", you mean "from an algebraic structure to itself".
 
  • #4
WWGD
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If [itex]x \in \ker h^2[/itex] then [itex]h(x) \in \ker h[/itex] (if [itex]x \in \ker h[/itex] then trivially [itex]h(x) \in \ker h[/itex]), so that [tex]\ker h^2 = h^{-1}(\ker h).[/tex] By the same logic [tex]\ker h^n = h^{-n+1}(\ker h).[/tex]

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.
 
  • #5
pasmith
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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 [itex]x \in \ker h^2[/itex] then by definition [tex]
x \mapsto h(x) \mapsto 0.
[/tex] Thus [itex]h(x) \in \ker h[/itex], ie. [itex]x \in h^{-1}(\ker h)[/itex]. Conversely, if [itex]x \in h^{-1}(\ker h)[/itex] then by definition [itex]h(x) \in \ker h[/itex], so that [itex]h^2(x) = 0[/itex], ie. [itex]x \in \ker h^2[/itex].

Consider the linear map [itex]h : \mathbb{R}^2 \to \mathbb{R}^2[/itex] with [itex]h(1,0) = 0[/itex] and [itex]h(0,1) = (1,0)[/itex]. Then [itex]\ker h[/itex] is the subspace spanned by (1,0) but [itex]\ker h^2 = h^{-1}(\ker h) = \mathbb{R}^2[/itex]: [tex]
(x,y) \mapsto (x,0) \mapsto (0,0).
[/tex]
 
  • #6
HallsofIvy
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You're right, but I also described the map as being from A to A.
I said it was a minor correction!
 
  • #7
WWGD
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It implies nothing of the sort.

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

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

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
 
  • #8
You have that ##h## maps ##\ker h^2## into ##\ker h##. This is not injective or subjective in general.
 

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