The Unapologetic Mathematician

Mathematics for the interested outsider

The First Isomorphism Theorem (for Abelian Categories)

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September 25, 2007 - Posted by | Category theory

8 Comments »

  1. […] Coimages, and Exactness By the first isomorphism theorem, we know that any morphism in an abelian category factorizes as with , and is epic. Since is […]

    Pingback by Images, Coimages, and Exactness « The Unapologetic Mathematician | September 26, 2007 | Reply

  2. […] We already said that (also ) is a abelian category. Now in any abelian category we have the first isomorphism theorem. […]

    Pingback by The Rank-Nullity Theorem « The Unapologetic Mathematician | June 27, 2008 | Reply

  3. I was wondering if there’s a way to do this in more general categories – for example, I was told about a “First Isomorphism Theorem for Sets” that goes as follows:

    In $\mathbf{Set}$, consider a map $f \colon X \to Y$, and consider $Z$ given by $X / \sim$ where $\sim$ is the equivalence relation given by $a \sim b$ if $f(a) = f(b)$.
    Then the First Isomorphism Theorem tells us that $\tilde{f} \colon Z \to \mathrm{im}(f)$ is an isomorphism. Is there anyway to put this in a more categorical setting?

    Comment by Sam | February 17, 2009 | Reply

  4. I’m sorry, it seems I misunderstood how LaTeX works on WordPress. The above post should have read:

    I was wondering if there’s a way to do this in more general categories – for example, I was told about a “First Isomorphism Theorem for Sets” that goes as follows:

    In \mathbf{Set}, consider a map f \colon X \to Y, and consider Z given by X / \sim where \sim is the equivalence relation given by a \sim b if f(a) = f(b).
    Then the First Isomorphism Theorem tells us that \tilde{f} \colon Z \to \mathrm{im}(f) is an isomorphism. Is there anyway to put this in a more categorical setting?

    Comment by Sam | February 17, 2009 | Reply

  5. Yes, if you consider carefully what properties of an Abelian category we’re using here. As I note above, this proof just uses the fact that we have a zero object, kernels, and cokernels.

    Now, \mathbf{Set} doesn’t have a zero object, so there are no kernels and cokernels. But there are equalizers and coequalizers, so that gives you something to start gnawing at.

    Comment by John Armstrong | February 17, 2009 | Reply

  6. […] it’s a very concrete representation of the first isomorphism theorem. Every transformation is decomposed into a projection, an isomorphism, and an inclusion. But here […]

    Pingback by The Meaning of the SVD « The Unapologetic Mathematician | August 18, 2009 | Reply

  7. […] for the singular value decomposition, what we’ll end up with is essentially captured in the first isomorphism theorem, but we’ll be able to be a lot more explicit about how to find the right bases to simplify […]

    Pingback by Decompositions Past and Future « The Unapologetic Mathematician | August 24, 2009 | Reply

  8. […] may not be Abelian, but it has a zero object, kernels, and cokernels, which is enough to get the first isomorphism theorem, just like for rings. Specifically, if is any homomorphism of Lie algebras then we can factor it […]

    Pingback by Isomorphism Theorems for Lie Algebras « The Unapologetic Mathematician | August 15, 2012 | Reply


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