1
$\begingroup$

I have the following map from the category of $k$-algebras to the category of sets:

$$\text{GL}_n:\mathbf{Alg}_k\rightarrow\mathbf{Set}:R\mapsto \text{GL}_n(R)$$

I would like to define its action on functions $f:R\rightarrow S$ to make it a functor.

Intuitively I would send $f:R\rightarrow S$ to $F:\text{GL}_n(R)\rightarrow \text{GL}_n(S):(a_{i,j})\mapsto (f(a_{i,j}))$

but nothing guarantees that the last matrix $(f(a_{i,j}))$ is indeed invertible.

Is there a way of showing that it is invertible? Or is it not the right way to define the action on functions?

$\endgroup$
  • $\begingroup$ Please provide more details! Your question is not clear at all! $\endgroup$ – Arman Malekzadeh Dec 30 '17 at 10:07
  • $\begingroup$ sorry I'll edit $\endgroup$ – tomak Dec 30 '17 at 10:10
2
$\begingroup$

Indeed this is a functor. It is usually thought of as a functor to the category of groups rather than sets, but never mind about that. The point is that if $A=(a_{ij})\in\text{GL}(R)$ then it has an inverse $B$, and $f(A)f(B)=f(AB)=f(I)=I$ where $f(A)=(f(a_{ij}))$, so that $f(A)$ has an inverse, as long as $f$ is a unital homomorphism, sending $1_R$ to $1_S$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.