"Evident" functor from CRng to Grp In _Categories for the Working Mathematician_ , p.14: > Functors arise naturally in algebra. To any commutative ring $K$, the set of all non-singular $n \times n$ matrices with entries in $K$ is the usual general linear group $GL_n(K)$; moreover, each homomorphism $f: K \rightarrow K'$ of rings produces in the evident way a homomorphism $GL_n f: GL_n(K) \rightarrow GL_n(K')$ of groups. The most evident functor for me would be applying $f$ to each coefficient of a matrix in $GL_n(K)$. But applying the zero homomorphism for example would yield the zero matrix, which is singular. What is the evident functor the author refers to ?

Your guess is correct. $GL_{n,f} $ applies $f$ to the entries. But this is not a problem, since the zero homomorphism is not a morphism in the category **Crng**. Namely each morphism $$g:R\to S$$ in this category, by definition fulfills $$ g(1_R)=1_S.$$ To see that $GL_{n,f}$ is well-defined, you can use the following:

1. An element $A\in GL_n(R)$ is invertible if and only if $det(A)\in R^*$
2. $det(GL_{n,f}(A))= f(det(A))$
3. Each ring homomorphism sends units to units



If you need more elaboration on any point, let me know.