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Difference between product and pullback

  1. Apr 4, 2012 #1
    What exactly is the difference between the categorical product of two objects and the fiber product (pullback) of two morphisms with the same two objects as domains? However I look at it, the morphisms in the definition appear superfluous.

    Can anyone display a category where products and pullbacks generally are not equal?

    Thanks in advance.
  2. jcsd
  3. Apr 4, 2012 #2


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    What is your definition of the product of two morphisms?
  4. Apr 4, 2012 #3
    I guess a reasonable definition is to do it in a slice category.
    When I work it out, I get the same diagram as with a pullback, but with two extra morphisms, but these are superfluous due to commutativity, so the pullback is the product of two objects in a slice category, or of two mophisms with a shared codomain.

    From wiki:

    But it seems that this forces the domain of the pullback to be the product of X and Y with p1 and p2 beign the usual projection morphisms. Is this not the case?

    Edit: I was mistaken. Of course, the pullback depends on the morphisms. For example in the category of abelian groups, if the two morphisms are trivial, the pullback is the trivial group, but if they are injective, the pullback is the regular product of the domains, correct?
    Last edited: Apr 4, 2012
  5. Apr 4, 2012 #4


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    No. If the morphisms are trivial (i.e. send everything to the identity) then the pullback is the regular direct product of the domains. If they are injective the pullback varies, depending on what the morphisms are. For example consider the pullback along id:G->G and id:G->G. As a set, it's ##\{(g_1,g_2) \colon g_1=g_2\}##, i.e., it's the diagonal in ##G\times G##.
  6. Apr 7, 2012 #5
    The pullback of f: X → Z and g: Y → Z is their product in the slice category of objects over Z.

    The product of X and Y is the pullback of X → 1 and Y → 1, where 1 is the terminal object.
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