Results 1 to 4 of 4

Math Help - meaning of a canonical map

  1. #1
    Newbie
    Joined
    Apr 2012
    From
    india
    Posts
    9

    meaning of a canonical map

    In my text, I often come across words like "there exists a canonical homomorphism between the group and the quotient group " for example -in the first isomorphism theorem . I do not understand what is meant by the word "canonical" in such contexts although I sort of get what is going on. May I request a clarification of the concept here? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: meaning of a canonical map

    "canonical" in this instance means "standard" (and in this instance is actually defined by a universal property).

    meaning: we do it in "the same way" no matter which particular group and normal subgroup we have.

    ok, let's look at it this way:

    suppose i have a group G, and a homomorphism f from G to some other group G'. suppose that H is a subgroup of G, and i say "f kills H". what might i mean by that?

    what i mean is: we have an obvious homomorphism of H into G defined by:

    iH(h) = h, for all h in H. and when i say f kills H, what i mean is:

    f∘i is the 0-homomorphism (that is: it maps everything to the identity e' of G'...the term "0" is somewhat mis-leading, we mean its a "trivial" homomorphism).

    now if N is a normal subgroup of G, we have a surjective homomorphism p:G-->G/N given by p(g) = gN (send every element of G to the coset of N it lives in).

    p is said to be "universal among homomorphisms that kill N". well, what does THAT mean?

    it means if we have a homomorphism f:G-->G', such that f(n) = e' for all n in N, f "factors through p", that is there is some OTHER homomophism f' with f = f'∘p.

    so let's unravel this. if f kills N, what we are saying is N is contained in ker(f), let's call ker(f), K (for kernel).

    what is this other homomorphism f'? it seems like it should be:

    f'(gN) = f(g), right? does this even make sense?

    well, first we need to check "well-defined-ness". that is we need to be sure that if gN = hN, that f(g) = f(h).

    if gN = hN, then h-1g is in N. but N is contained in K, so h-1g is also in K, which means f(h-1g) = e' (since K is the kernel of f).

    since f is a homomorphism, f(h-1g) = f(h-1)f(g) = [f(h)]-1f(g). since this also equals e', we have:

    f(g) = f(h) (multiply by f(h) on the left). so f' is indeed at least a function from G/N to G'.

    to see that f' is actually a homomorphism, we compute:

    f'((gN)(hN)) = f'((gh)N) = f(gh) = f(g)f(h) = f'(gN)f'(hN).

    finally we check that f'∘p = f:

    f'∘p(g) = f'(p(g)) = f'(gN) = f(g).

    this is all pretty much the same thing you see in the first isomorphism theorem, it's not all THAT exciting. so why bother?

    well...because proving things about N as a subgroup of G, and about G/N as a factor group (quotient group) usually depends on (like we did above) picking "typical elements" g,h in G and seeing what happens to them.

    but saying: p is universal among homomorphisms that kill N isn't a statement about ELEMENTS of G, its a statement about our "canonical" homomorphism p. in other words, instead of looking at the OBJECT G/N, we're looking at the MAPPING p.

    and we're saying: "if a homomorphism (any homomorphism) kills N, it has to "go through p" first. p is "special" or "distinguished" among all the homomorphisms that kill N, it's OPTIMAL (it kills N, and ONLY N, not anything more).

    that is: given the problem of a group G, and a normal subgroup N, how do we make a homomorphism that "just kills N"? form the factor group (apply the homomorphism p) G/N.

    this construction occurs so frequently, we don't even bother giving p "a name", because we do it "the same way" for every group G and any normal subgroup N.

    in this light, what the first isomorphism theroem says is:

    there is no practical difference between a factor group G/N and a homomorphic image of G. we can use either "construction" as it suits us. sometimes the quotient group G/N is easier to understand, sometimes f(G) is easier to deal with.

    so rather than carry the letter p around and explicitly define it (for a particular subgroup N, and a particular group G), we just say:

    consider the canonical homomorphism G-->G/N, which you are expected to know means "p".
    Last edited by Deveno; December 30th 2012 at 04:52 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2012
    From
    india
    Posts
    9

    Re: meaning of a canonical map

    Thanks a lot Deveno for clearing my doubt.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: meaning of a canonical map

    Hi clerk!

    The word canonical means obvious.

    The canonical homomorphism of Z \to Z/3Z is k \mapsto \bar k.
    This is basically just the identity, but we can't call it that because it is not really the identity, so we call it the canonical mapping.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Canonical representation?
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: January 16th 2012, 08:16 AM
  2. canonical parameter and canonical link function
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 6th 2011, 08:41 AM
  3. Doubts about the meaning of "canonical"
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 28th 2010, 02:45 PM
  4. Canonical coordinates
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 15th 2010, 11:09 AM
  5. Canonical Basis?
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: June 6th 2007, 10:46 PM

Search Tags


/mathhelpforum @mathhelpforum