# Theorem dual to trivial kernel $\iff$ injective in $\boldsymbol{Grp}$

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?)

• 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. May 13, 2020 at 20:23
• @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 $S_3$
– user736690
May 13, 2020 at 20:23
• 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. May 13, 2020 at 20:25
• The dual theorem of course exists. But it applies to the opposite category $\mathbf{Grp}^{op}$, which is not equivalent to $\mathbf{Grp}$. May 13, 2020 at 21:03

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 $$\mathcal (C^{op})^{op} = \mathcal C$$.
• 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. May 13, 2020 at 22:27