Fibre-discrete concrete categories.
The question is in the process of proving the statement in “Abstracte and Concrete Categories” book from the $\mathbf E\mathbf x. 5\mathbf E (a)$ on the page 78. The first and the second statements.
They are here. I linked the book to help others to use its definitions.
Show that no proper subconstruct of $\mathbf G \mathbf r\mathbf p$ is concretely reflective (or coreflective). Generalize this to all fibre-discrete concrete categories.
As for me, the only way to find the proof is to come to contradiction, but I do not have any idea to find them. Of course, we need to use somehow the knowledge that this category is fibre-discrete but through the concrete reflector it remains to be fibre-discrete. Does someone know how to prove this statement?
Thanks.
$\endgroup$1 Answer
$\begingroup$By definition, a subconstruct $A$ of $\mathbf{Grp}$ is concretely reflective if and only if for every group $G$ there exist an "identity-carried $A$-reflection", i.e. an identity-carried group homomorphism $\varphi:G\to H$ such that $H$ is an object of $A$, satisfying a certain universal property. Here "identity-carried" means that the forgetful functor maps $\varphi$ to an identity; but this implies that $G$ and $H$ must have the same underlying set, and be endowed with the same group structure, so $G=H$. Thus in fact every group $G$ must be in $A$, which means that $A$ cannot be proper. The proof in fibre-discrete categories is exactly the same : being fibre-discrete means that the only identity-carried morphisms are identities.
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