# Why does the group of units functor respect products? That's I believe we have $(X×Y)^×\cong X^××Y^×$, in general. Why?

Perhaps someone can explain this category theoretic fact. I must have read somewhere that it follows from the fact that there's a left adjoint for this functor. That would be the group ring construction.

So I guess my question boils down to: why does the existence of a left adjoint for a functor guarantee that it respects (preserves) products?

• if a functor has a left adjoint then that functor itself is a right adjoint, and "right adjoints preserve limits" is a standard result, e.g. math.stackexchange.com/questions/101005/… or ncatlab.org/nlab/show/adjoints+preserve+%28co-%29limits or maths-magic.ac.uk/downloads/course-file/6255 (and of course a product is a type of limit) Commented Feb 8, 2021 at 14:22
• Thanks @MatthewTowers So it is a standard result. I figured as much.
– user403337
Commented Feb 8, 2021 at 14:26
• @MatthewTowers Could you point out how a product is a type of limit?
– user403337
Commented Feb 8, 2021 at 14:37
• the product of X and Y is the limit of the functor from the two object discrete category which sends one object to X and the other to Y. Commented Feb 8, 2021 at 15:17
• Okay @MatthewTowers My category theory is quite weak. If Stallings were still alive, and I had the nerve, I might ask him about "those functor things".
– user403337
Commented Feb 8, 2021 at 15:20

a unit in $$X\times Y$$ is a couple $$(x,y)$$ with an inverse, that is, a couple $$(x',y')$$ with $$(x,y)(x',y') = (1,1)$$. Since multiplication is defined pointwise, this is equivalent to $$xx' = 1, yy' = 1$$ (and symmetrically for $$x'x, y'y$$) so that if $$x,y$$ are invertible, so is $$(x,y)$$.
Now here's the category theoretic version (this is a summary from the comments essentially): $$R\mapsto R^\times$$ is right adjoint to $$G\mapsto \mathbb Z[G]$$ as is easily checked, so that it must commute with all limits : this is a general fact, right adjoints always preserve limits.
Moreover, products are limits: $$A\times B$$ is the limit of the functor from the category $$\bullet \space\bullet$$ with values $$A$$ and $$B$$
The proof in that particular case could go as follows: $$\hom(G,(X\times Y)^\times) \cong\hom(\mathbb Z[G],X\times Y)\cong \hom(\mathbb Z[G],X)\times \hom(\mathbb Z[G],Y)\cong \hom(G,X^\times)\times \hom(G,Y^\times) \cong \hom(G,X^\times\times Y^\times)$$ and so $$(X\times Y)^\times \cong X^\times \times Y^\times$$ by the Yoneda lemma.