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I was looking an example of determinant as natural transformation. I have tried to prove it, but my question here is regarding the categories in which we are looking it as a functor.

Fix a natural number $n$. Let Comm denotes the category of commutative rings. For every object $R$ in Comm, we have a new ring $M_n(R)$. Let Mon denotes the category of monoids. Then $M_n(R)$ is an object in the Mon for every object $R$ in Comm.

Now $\mathcal{F}:$ Comm $\rightarrow$ Mon, $R\mapsto M_n(R)$.

Let $\mathcal{G}:$ Comm $\rightarrow$ Mon, $R\mapsto R$ (where $R$ in domain is commutative ring, and $R$ in co-domain is monoid w.r.t. multiplication).

It is then proved that determinant is a natural transformation from $\mathcal{F}$ to $\mathcal{G}$.

I have proved this using the Axiometic definition of natural transformation (I mean, beyond Axiometic verification, I have not understood much).

But, my question here is the following. The objects $M_n(R)$ considered above belong to a category of monoids; but at the same time, they are also rings, and so we can consider them as objects in the category Rings of rings. Is there any specific reason to consider $M_n(R)$ as objects in Mon instead of Rings? In other words,

What problem arises if we consider $\mathcal{F}$ and $\mathcal{G}$ as functors from Comm to Rings?

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  • $\begingroup$ A book describing this approach is Leinster: Basic Category Theory, page 29 $\endgroup$ – Javier Apr 17 at 12:54
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The problem is that $\det$ is only a multiplicative monoid homomorphism, not a ring homomorphism. That is, $\det(A+B)\neq\det(A)+\det(B)$ in general. So if you considered your functors as taking values in $\mathbf{Rings}$, $\det$ would not even be a morphism in that category for any fixed $R$.

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  • $\begingroup$ Eric, thanks for the clarification. $\endgroup$ – p Groups Sep 27 '16 at 8:01

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