Cokernels are "dual" to kernels. I've been told that if X dual to Y and there's a theorem about X then there's a "dual" theorem about Y.

Theorem: In $\boldsymbol{Grp}$ a homomorphism has trivial kernel iff it is injective.

However it is not true that a group homomorphism has trivial cokernel iff it's surjective. So what would instead be the dual theorem to the above theorem? (or have I been lied to and those dual theorems don't always exist?)

  • 1
    $\begingroup$ Cokernels in $\mathrm{Grp}$ are not really cokernels in the strict sense: they do not fulfill the universal property of cokernels, which the failure of your dualization is a witness of. If you work in an Abelian category (such as the category of Abelian groups) instead, this dualization works, and your proposition is true. $\endgroup$ May 13, 2020 at 20:23
  • $\begingroup$ @BenSteffan book i'm reading claims that there are non-surjective group homomorphisms with trivial cokernel. They give as an example the inclusion map of the subgroup <(12)> of $S_3$ $\endgroup$
    – user736690
    May 13, 2020 at 20:23
  • $\begingroup$ I realized that your definition defines the cokernel of a map as the quotient by the normal closure of its image, not as the quotient by the image itself (which is only a group if the image is normal), which is why I was confused. $\endgroup$ May 13, 2020 at 20:25
  • $\begingroup$ The dual theorem of course exists. But it applies to the opposite category $\mathbf{Grp}^{op}$, which is not equivalent to $\mathbf{Grp}$. $\endgroup$ May 13, 2020 at 21:03

1 Answer 1


The dual theorem holds in the opposite category $\boldsymbol{Grp}^{op}$, which is not equivalent to $\boldsymbol{Grp}$.

This sort of duality holds in general when a theorem is true for all categories, or at least some subset of categories that are self dual. If for all categories $\mathcal C$, some theorem holds for $\mathcal C$, then the dual theorem holds for $\mathcal C^{op}$. Since the original theorem also holds for the category $\mathcal C^{op}$, the dual theorem holds for $\mathcal (C^{op})^{op} = \mathcal C$.

This particular theorem doesn't hold for all categories (where kernels, cokernels, injectivity and surjectivity can be defined). Thus, the dual theorem is not guaranteed to hold.

  • $\begingroup$ How does that category look like? Is there an easy description for it? $\endgroup$
    – user736690
    May 13, 2020 at 21:28
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    $\begingroup$ That might warrant another question along the lines of this one or this one. If pressed, I'd use the answer to that first question to say that $\boldsymbol {Grp}^{op}$ is the category of cogroup objects in the category of complete atomic boolean algebras. $\endgroup$
    – SCappella
    May 13, 2020 at 22:27

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