# Fibre-discrete concrete categories

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?

Thanks.

Let $\mathbf{A}$ be a concretely reflective subconstruct of $\mathbf{C}$, a fibre-discrete category.
Let $C\in \mathbf{C}$ any object and $C\to^f A$ be an identity carried $\mathbf{A}$-reflection arrow.
But then, in $\mathbf{C}$, $C\leq A$ since $f$ is identity-carried. But $\mathbf{C}$ being fibre-discrete, its fibers are ordered by equality, so $A=C$, and so $C\in \mathbf{A}$ : $Ob(\mathbf{A}) = Ob(\mathbf{C})$. Let moreover $g: C\to D$ be a $\mathbf{C}$-morphism. Since $D$ is an $\mathbf{A}$-object, we can use the reflection property to get an $\mathbf{A}$-arrow $j: A\to D$ such that the appropriate diagram commutes. But then since $C\to^f A$ is the identity, $j=g$ and so $g$ is an $\mathbf{A}$-arrow : $\mathbf{A}=\mathbf{C}$.
The construct $\mathbf{Grp}$ (and indeed any construct of the form $\mathbf{Alg}(\Omega)$) is obviously fibre-discrete and so this applies.
• I’m not sure,but the fact that $A=C$ means that there exist morphisms $f:C\rightarrow A$ and $g:A\rightarrow C$, so can it be in such way that $C\notin \mathbf A$? – A. Gonus Oct 3 '17 at 21:13
• No $A=C$ is an equality, not an isomorphism. A category being fibre-discrete means its fibres are ordered by equality: $A\leq C \implies A=C$ – Maxime Ramzi Oct 3 '17 at 21:22