The question is in the process of proving the statement in “Abstracte and Concrete Categories” book http://katmat.math.uni-bremen.de/acc/acc.pdf 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?



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.

  • $\begingroup$ But if “identity-carried” means in such constructions (i.e., subconstructs in concrete categories) that the Reflector $\mathbf R:\mathbf B\rightarrow \mathbf A$, where $\mathbf B$ is a concrete category and $\mathbf A$ is its subconstruct, maps $\phi$ to an identity? Or we are always assume this property for the functor from concrete category pair $(\mathbf A,U)$? $\endgroup$ – A. Gonus Oct 3 '17 at 12:16
  • $\begingroup$ How do I have to distinguish these two cases? @Arnaud $\endgroup$ – A. Gonus Oct 3 '17 at 13:23
  • $\begingroup$ No, identity-carried means that the forgetful functor $U:\mathbf{B}\to \mathbf{A}$ carries $\varphi$ to $id$, not the reflector. If you wanted the reflector $R$ to carry $\phi$ to an identity, then the subcategory of abelian groups would be concretely reflective, but this is not the case (see Examples 5.23 (4) ) $\endgroup$ – Arnaud D. Oct 3 '17 at 13:25
  • $\begingroup$ Did you mean the forgetful functor $U:\mathbf B\rightarrow\mathbf S\mathbf e\mathbf t$ @Arnaud? $\endgroup$ – A. Gonus Oct 3 '17 at 13:40
  • $\begingroup$ Yes, sorry about the confustion. $\endgroup$ – Arnaud D. Oct 3 '17 at 13:40

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