After a re-read, two questions to see if I have this straight:
Given φ: X→Y where A⊆X and B⊆Y, is it fair to state: the image of A is crudely speaking just φ[A] = B, and the preimage of B is likewise just φ⁻¹[B] = A?
Is a kernel then essentially the preimage of {e'}?
If indeed φ[A] = B, then B is the image of A, yes. And, if A is the set containing ALL elements that get sent to B, then φ⁻¹[B] = A.
To make sure the difference is clear, let's use a really goofy example. Let φ : ℤ -> {0} be defined in the only way that makes sense, that is, φ(n) = 0 for any n. Then if A = {-1, 0, 1}, we have φ[A] = B, but NOT φ⁻¹[B] = A, because φ⁻¹[B] = ℤ. Does that make sense? A may "cover" B without exhausting everything that goes to B.
And yes, the kernel of φ is precisely the preimage of {e'}!
Okay, cool. Glad you clarified it isn't always reciprocal. I wanted to be sure I understood that the definition, φ[A] := {φ(a) | a ∈ A}, effectively meant that φ[A] = B. Likewise, that the definition, φ⁻¹[B] := {a ∈ A | φ(a) ∈ B}, meant φ⁻¹[B] = A. And now I understand, with A and B suitably defined, possibly differently, in each case.
Still struggling with the first section of VII. I'm going to have to construct some groups and work out some examples. Back at it tomorrow!
>> "What remains to be shown is closure (since associativity is inherited from G'.) Let a, b ∈ H. We know that φ(a) and φ(b) are in φ[H]. What we need to show is that φ(a)φ(b) ∈ H."
Had to go back and review to tackle the current lesson. Should the above end: "...φ(a)φ(b) ∈ φ[H]"? The text continues, "Because φ is a homomorphism, we have φ(ab) = φ(a)φ(b), so φ[H] indeed contains φ(ab)." Which is what made me think so. (I'm apparently finding the concept of a kernel hard to wrap my head around.)
I'm going to have to chew on this some more before I try the exercises. I'm not clear on a couple of things here:
What is the difference between φ[A] and φ(A)? It seems as if they would give the same result.
Proving the first item of the first theorem, I'm not sure I understand the algebra. Multiplying both sides of φ(e) = φ(e)φ(e) by φ⁻¹(e), I get φ(e)/φ(e) = φ(e)φ(e)/φ(e), and I don't follow why the left side cancels to e' while the φ(e)/φ(e) on the right just goes away. I get that e^2=e, and I can see e/e=e, but I'm not clear on exactly where the e' came from.
I think the reason this notation is used for φ[A] and φ(A) is to differentiate when we're thinking about an individual object and a set.
The fractions both cancel to e', on the left and on the right. So we get φ(e)/φ(e) = φ(e)φ(e)/φ(e), but φ(e)/φ(e) = e' (by the definition of the inverse - remember that φ(e) is an object in G'). So the LHS is e', and the RHS is φ(e) • e' = φ(e). Does that clear things up?
Thanks, yes. I had a feeling φ[A] versus φ(A) might have to do with the set versus its members. I did end up with e' = φ(e) • e', but it somehow ended up feeling more like a tautology than a proof. Maybe because φ(e) = e' is given by the homomorphism?
After a re-read, two questions to see if I have this straight:
Given φ: X→Y where A⊆X and B⊆Y, is it fair to state: the image of A is crudely speaking just φ[A] = B, and the preimage of B is likewise just φ⁻¹[B] = A?
Is a kernel then essentially the preimage of {e'}?
If indeed φ[A] = B, then B is the image of A, yes. And, if A is the set containing ALL elements that get sent to B, then φ⁻¹[B] = A.
To make sure the difference is clear, let's use a really goofy example. Let φ : ℤ -> {0} be defined in the only way that makes sense, that is, φ(n) = 0 for any n. Then if A = {-1, 0, 1}, we have φ[A] = B, but NOT φ⁻¹[B] = A, because φ⁻¹[B] = ℤ. Does that make sense? A may "cover" B without exhausting everything that goes to B.
And yes, the kernel of φ is precisely the preimage of {e'}!
Okay, cool. Glad you clarified it isn't always reciprocal. I wanted to be sure I understood that the definition, φ[A] := {φ(a) | a ∈ A}, effectively meant that φ[A] = B. Likewise, that the definition, φ⁻¹[B] := {a ∈ A | φ(a) ∈ B}, meant φ⁻¹[B] = A. And now I understand, with A and B suitably defined, possibly differently, in each case.
Still struggling with the first section of VII. I'm going to have to construct some groups and work out some examples. Back at it tomorrow!
>> "What remains to be shown is closure (since associativity is inherited from G'.) Let a, b ∈ H. We know that φ(a) and φ(b) are in φ[H]. What we need to show is that φ(a)φ(b) ∈ H."
Had to go back and review to tackle the current lesson. Should the above end: "...φ(a)φ(b) ∈ φ[H]"? The text continues, "Because φ is a homomorphism, we have φ(ab) = φ(a)φ(b), so φ[H] indeed contains φ(ab)." Which is what made me think so. (I'm apparently finding the concept of a kernel hard to wrap my head around.)
Yes, that’s correct. I’ll go back and fix this once I’m back at my computer. Thanks for catching that!
I'm going to have to chew on this some more before I try the exercises. I'm not clear on a couple of things here:
What is the difference between φ[A] and φ(A)? It seems as if they would give the same result.
Proving the first item of the first theorem, I'm not sure I understand the algebra. Multiplying both sides of φ(e) = φ(e)φ(e) by φ⁻¹(e), I get φ(e)/φ(e) = φ(e)φ(e)/φ(e), and I don't follow why the left side cancels to e' while the φ(e)/φ(e) on the right just goes away. I get that e^2=e, and I can see e/e=e, but I'm not clear on exactly where the e' came from.
I think the reason this notation is used for φ[A] and φ(A) is to differentiate when we're thinking about an individual object and a set.
The fractions both cancel to e', on the left and on the right. So we get φ(e)/φ(e) = φ(e)φ(e)/φ(e), but φ(e)/φ(e) = e' (by the definition of the inverse - remember that φ(e) is an object in G'). So the LHS is e', and the RHS is φ(e) • e' = φ(e). Does that clear things up?
Thanks, yes. I had a feeling φ[A] versus φ(A) might have to do with the set versus its members. I did end up with e' = φ(e) • e', but it somehow ended up feeling more like a tautology than a proof. Maybe because φ(e) = e' is given by the homomorphism?