If $\mathcal{C}$ is a category, $A\in\mathcal{C}$ is an abelian object if $\mathcal{C}(-,A)$ is naturally an abelian group. Assuming $\mathcal{C}$ has enough limits, this is equivalent to there being a multiplication morphism $m:A\times A\to A$, an identity $\epsilon:*\to A$ and an inverse $i:A\to A$, such that all the diagrams commutes.

I take this to mean $C(T,A)$ is an abelian group for every $T\in\text{Ob}(\mathcal{C})$. Is that the correct interpretation? How then does 'having enough limits' induce this?

(Secondary curiosity, if anyone knows, would this mean that the $H$-spaces are the abelian objects in the category of pointed topological spaces, with morphisms being homotopy classes of maps.)

  • $\begingroup$ see ncatlab.org/nlab/show/group+object $\endgroup$ – Josselin Poiret Jul 30 '18 at 10:19
  • $\begingroup$ @FreeSalad Apparently it reduces to that (somehow) when $\mathcal{C}$ has enough limits. But I imagine these are distinct concepts otherwise? $\endgroup$ – user580497 Jul 30 '18 at 10:21
  • $\begingroup$ well, the question here is that we need a cartesian monoidal structure on a category to state the group axioms, so for categories which are of that nature it is easy to define internal groups, however for all other categories we have access to its presheaf category which is cartesian closed itself. It is just a matter of lifting the definition to the presheaf category in a compatible way. $\endgroup$ – Josselin Poiret Jul 30 '18 at 10:55

The part "$\mathcal C(-,A)$ is a naturally abelian group" means that there exists a functor $\mathbf A : \mathcal C^{\rm op} \to \mathsf{Ab}$ such that that $U \circ \mathbf A$ and $\mathcal C(-,A)$ are isomorphic (as functors $\mathcal C^{\rm op} \to \mathsf{Set}$) where $U$ is the forgetful functor $\mathsf{Ab} \to \mathsf{Set}$.

The part "$\mathcal C$ has enough limits" is just to justify the existence of the product $A\times A$ and of the terminal object $\ast$ in use just after.

Now that the vocabulary is set, the proof is more or less tautological. Given maps $\epsilon, i, m$ making the wanted diagrams commute, you can defined a "pointwise" abelian group structure on $\mathcal C(X,A)$: the unit is $X\to \ast\overset \epsilon\to A$, the inverse operation is taking $f: X \to A$ to $X\overset f\to A \overset i\to A$, and the multiplication takes $f,g: X\to A$ to $X\overset\delta\to X\times X \overset {f\times g}\to A\times A \overset m \to A$. You need to check that it is indeed an abelian group, and that it is functorial.

Conversely, given that "$\mathcal C(-,A)$ is naturally an abelian group", you get natural transformations $$\boldsymbol\epsilon :\mathcal C(-,\ast) \to \mathcal C(-,A) \\ \mathbf i:\mathcal C(-,A) \to \mathcal C(-,A) \\ \mathbf m : \mathcal C(-,A\times A)\simeq \mathcal C(-,A)\times\mathcal C(-,A) \to \mathcal C(-,A)$$ defined at component $X$ as the unit, the inverse and the multiplication of $\mathbf A(X)$ respectively. You need to check that this is actually natural. By Yoneda lemma, it gives $\epsilon, i, m$ such as wanted. The fact that they make all the wanted diagrams commute will be given by $\mathbf A(X)$ being an abelian group for each $X$.


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