Prove the determinant map is a natural transformation I'm trying to learn about natural transformations, but I'm lost trying to work through the details of one.
Problem: Let $Comm$ be the category whose objects are commutative rings and whose maps are ring homomorphisms, and let $M$ be the category whose objects are monoids and maps are monoid homomorphisms.  Then fix $n \in \mathbb{N}$ and let $F: Comm \to M$ be the functor defined by $F(R)=M_{n}(R)$ and $G: Comm \to M$ be the functor $G(R)=R^{\star}$ (the ring $R$ under multiplication).  Interpret the determinant map as a natural transformation.
Attempted proof.  Since determinant is a function from a matrix to the elements of it corresponding ring, the proposed natural transformation is a family of maps in $M$ where for any matrix monoid $M_{n}(R)$ there is the determinant map to the multiplicative monoid $R^{\star}$.  Since $\det(I)=1$ and $\det(AB)=\det(A)\det(B)$, we know that $\det$ is a homomorphism so there actually is a determinant map for any matrix monoid in $M$.  To show this is a natural transformation, we need to shows that it commutes, so we fix a map $f: A \to B$ in $Comm$ and we need to show $\det \circ F(f)=G(f) \circ \det$.  We know $F(f)$ is a map from $M_{n}(A)$ to $M_{n}(B)$ and thus $\det \circ F(f)$ is a map from $M_{n}(A)$ to $B^{\star}$.  Similarly, $G(f)$ is a map from $A^{\star}$ to $B^{\star}$ and det maps $M_{n}(A)$ to $A^{\star}$ so $G(f) \circ \det$ is also a map from $M_{n}(A) \to B^{\star}$.  
This is the point where I get a little lost.  It seems to me that the way to show these two maps are equal is to fix some $X\in M_{n}(A)$ and show that both homomorphisms send $X$ to the same element of $B^{\star}$.  For the right hand side, we have $G(f) \circ \det (A)=G(f) (\det( A))=f(\det(A))$.  This looks confusing to me, but if I understand correctly, the functor $G$ sends a map/homomorphism $f$ in $Comm$ to the identical map/homomorphism in $M$ and the real "heart" of $G$ is how it forgets the additive structure of objects in $Comm$.  Do I understand this correctly?  Then, for the other map, we want to look at $\det \circ F(f) (A)$, but I don't really understand what $F(f)(A)$ is.  We know $F(f)$ is some homomorphism from $M_{n}(A)$ to $M_{n}(B)$, but I don't know what homomorphism it actually is, so I'm stuck.
 A: The point is that determinant is essentially given by the same algebraic formula for all rings, and ring homomorphisms respect it. 
For instance, consider $n = 2$, so that $F(R) = M_2(R)$. Fix some ring homomorphism $f: A \to B$. Now let's show that $\det_B \circ F(f) = G(f) \circ \det_A$, where $\det_R: M_2(R) \to R^\star$ is obviously $\det([[a, b], [c, d]]) = ad - bc$. 
Then, to show $\det_B \circ F(f) = G(f) \circ \det_A$ we need to show that for any $M \in M_2(R)$ it holds that $\det_B(F(f)(M)) = G(f)(\det_A(M))$. 
Now, let $M = [[a, b], [c, d]]$. Then $F(f)(M) = [[f(a), f(b)], [f(c), f(d)]]$, and so $\det_B(F(f)(M)) = f(a)f(d) - f(b)f(c) = f(ad-bc)$, and the last equality follows from the fact that $f$ is ring homomorphism.
On the other hand, $G(f)(\det_A(M)) = G(f)(ad - bc) = f(ad-bc)$, since $G(f)$ is the same map as $f$, set-theoretically. Thus, we obtain the desired equality. 
The case for $n>2$ is similar, as you'll clearly see.
A: First of all, your discussion of the functor $G$ looks good.
The objects in the category $Comm$ are commutative rings, so a morphism $f: A \to B$ is a unital ring homomorphism.  The functor $F$ maps a ring to the monoid of matrices over that ring.  How does this functor act on morphisms?  Entry-wise!  Given $X = \left( x_{ij} \right) \in M_n(A) = F(A)$,
$$
F(f): M_n(A) \to M_n(B)
$$
maps
$$
\left( x_{ij} \right) \mapsto \left( f(x_{ij}) \right).
$$

By the way, in my opinion, the multiplicative monoid is not necessarily the "heart" of a ring, as there are really two forgetful functors from $Comm$.  The first is your $F$ that just keeps around the multiplicative monoid.  The other just keeps around the abelian group.
Similar exercise:  Is it true that with the forgetful functor $F':  Comm \to Ab$, the trace map is a natural transformation?
A: Let me make a precision to what have been said. 
As said before, $F(f)$ can be taken to be the application of $f$ to each of the components of the respective matrix and $G(f)$ is essentially the same $f$ but formally working on a monoid instead of a ring.
If you want to proof rigorously that the induced mappings $F(f)$ and $G(f)$ conmute with the determinant then just take the general expression of the determinant in a commutative ring, Wikipedia calls it Leibnitz's formula: $$\det(A) = \sum_{\sigma \in S_n} (sgn(\sigma) \prod_{i = 1}^n a_{i,\sigma_i})$$
You will admit that since ring homomorphism $f$ preserves sums and products, the conclusion is straight forward.
Reference
Tom Leinster, Basic Category Theory, page 30
