Let $R$ be any commutative ring with $1$. Then every homomorphism from $R^n$ to $R^n$ has a unique matrix representation. This means we have a $R$-algebra(?) isomorphism $$\mathrm{Hom}(R^n,R^n)\to R^{n\times n}\cong R^n\otimes R^n$$ although the multiplication in $\mathrm{Hom}(R^n,R^n)$ is not commutative. I didn't learn anything about non-commutative $R$-algebras, so I don't know if this identification is right. Anyway $\mathrm{det}$ is a multiplicative map from $\mathrm{Hom}(R^n,R^n)$ to $R$ satisfying $\det(AB)=\det(A)\det(B)$. The problem is that this is not an $R$-algebra homomorphism, or an $R$-module homomorphism, or even not a ring homomorphism. This is even not a group homomorphism since $\det(A)$ may not be a unit in $R$. Maybe at least I can say that this is a monoid homomorphism from $\mathrm{Hom}(R^n,R^n)$ to $(R,\times)$, but I don't think this is a good description for the determinant.
So what is the determinant? Is there any categorical way to define the determinant map?
Thanks in advance.
I found the answer myself here https://ncatlab.org/nlab/show/determinant+line+bundle.
When a $k$-linear map $T:V\to W$ over $k$-vector spaces is given with $\dim V= \dim W =n$, this induces a $k$-linear map $$ \det T:\bigwedge^nV \to \bigwedge^nW $$ $\bigwedge^nV$ is isomorphic to $k$, so $\det T(1)$ is exactly the original definition of $\det T$. The multiplcation law $\det(AB)=\det(A)\det(B)$ is just the functoriality of $\bigwedge^n$.