A functor is amnestic if its domain has no more duplication of isomorphic objects than its codomain. This formalizes the sense in which (for example) the concrete category $Met_cont$ of metric spaces and continuous maps is often better thought of as the category $Met Top$ of metrizable topological spaces (and continuous maps), by identifying a property that the forgetful functor $Met Top \to Set$ has but $Met_cont \to Set$ does not. As ‘amnestic’ is basically a fancy synonym of ‘forgetful’, the idea is to identify a property that one would like in a forgetful functor.
There is a corresponding notion of amnesticization of a functor (called the amnestic modification in The Joy of Cats), which replaces the domain with an equivalent category, relative to which the functor becomes amnestic. Applying this to $Met_cont \to Set$ produces a category isomorphic to $Met Top$. Although not needed in this case, we need the axiom of choice (AC) in general to prove that every functor has an amnesticization. Without AC, we can use amnestic anafunctors to make everything work out, although much of the convenience is lost.
Amnesticity is really a property of strict functors (or anafunctors) between strict groupoids. Groupoids, because the non-isic morphisms play no role in the definition; only the categories' cores matter, and a functor is amnestic iff its core is amnestic. Strict, because the definition requires us to state (in two places) that some isomorphic objects are equal; weakening the definition to follow the principle of equivalence leads to a trivial property that every functor satisfies. (That is, up to equivalence, every functor is amnestic, which is because every functor is equivalent to its amnesticization.)
Let $C$ and $D$ be two strict categories, and let $U$ be a strict functor from $D$ to $C$. We say that $U$ is amnestic if its core reflects identity morphisms. Explicitly, $U$ is amnestic iff, whenever $a$ and $b$ are objects of $D$, $f$ is an isomorphism in $D$ from $a$ to $b$, $U(a)$ and $U(b)$ are equal objects in $C$, and $U(f)$ is the identity morphism on this object in $C$, then $a$ and $b$ are equal objects in $D$, and $f$ is the identity morphism on this object in $D$.
If we follow the principle of equivalence and refuse to state equations between objects, then we must modify the hypothesis to say that $U(a)$ and $U(b)$ are isomorphic in $C$ (say via $g\colon U(a) \to U b$) and $U(f)$ is the identity relative to this isomorphism (so $U(f) = \id_{U(b)} \circ g \circ \id_{U(a)}$; since we can simply let $g$ be $U(f)$, this is trivial (beyond the initial isomorphism $f\colon a \to b$). Similarly, we must modify the conclusion to say that $a$ and $b$ are isomorphic (say via $h\colon a \to b$) and $f$ is the identity relative to this isomorphism (so $f = \id_b \circ h \circ \id_a$); since we can simply let $h$ be $f$, this is also trivial. Thus up to equivalence, this property is trivial; on the other hand, it is preserved by isomorphism.
Now let $C$ and $D$ be two strict categories, and let $U$ be an anafunctor from $D$ to $C$. We again say that $U$ is amnestic if its core reflects identity morphisms. Explicitly, now $U$ is amnestic iff, whenever $a$ and $b$ are objects of $D$, $f$ is an isomorphism in $D$ from $a$ to $b$, $\alpha$ is a specification of $a$ for the anafunctor $U$, $\beta$ is a specification of $b$ for the anafunctor $U$, $U_\alpha(a)$ and $U_\beta(b)$ are equal objects in $C$, and $U_{\alpha,\beta}(f)$ is the identity morphism on this object in $C$, then $a$ and $b$ are equal objects in $D$, and $f$ is the identity morphism on this object in $D$.
We might also demand that $\alpha = \beta$; this is automatic if $U$ is saturated.
An amnestic full and faithful functor is automatically an isocofibration, i.e. injective on objects: if $U D' = U D$, then there is some isomorphism $f : D' \to D$ in $\mathcal{D}$ such that $U f = id_{U D}$, but then we must have $f = id_{U D'} = id_{U D}$, so $D' = D$.
An amnestic isofibration has the following lifting property: for any object $D$ in $\mathcal{D}$ and any isomorphism $f : C \to U D$ in $\mathcal{C}$, there is a unique isomorphism $\tilde{f} : \tilde{C} \to D$ such that $U \tilde{f} = f$. Indeed, if $\tilde{f}' : \tilde{C}' \to D$ were any other isomorphism such that $U \tilde{f}' = f$, then $U (\tilde{f}^{-1} \circ \tilde{f}') = id_C$, so we must have $\tilde{f} = \tilde{f}'$.
If the composite $U \circ K$ is an amnestic functor, then $K$ is also amnestic.
Most famous forgetful functors are amnestic, such as $Grp \to Set$, $Top \to Set$, $Top Grp \to Grp$, and $Top Grp \to Top$. Even $Met \to Set$ is amnestic, since the morphisms in Met are short maps. However, $Met_cont \to Set$, where the morphisms are continuous maps, is not amnestic, as is shown by any set with two different but topologically equivalent metrics (such as $\mathbb{R}^2$ with the $l^1$ and $l^\infty$ metrics).
The forgetful functor from a groupoid of structured sets is amnestic. The examples above may all be defined by starting from such a groupoid and specifying which functions preserve the structure (and so are morphisms in the category of structured sets). So long as this produces no additional isomorphisms, the forgetful functor will be amnestic. But in the case of $Met_cont$, any non-isometric homeomorphism will be an isomorphism that was not in the original groupoid (consisting only of surjective isometries), and so this forgetful functor is not amnestic.
Any strictly monadic functor is amnestic. Conversely, any monadic functor that is also an amnestic isofibration is necessarily strictly monadic.
J. Adámek, H. Herrlich and G.E. Strecker: The Joy of Cats, Chapter I, Definition 3.27.
G. Preuß: Theory of Topological Structures: An Approach to Categorical Topology. D. Reidel Publishing Company. Mathematics and Its Applications, Dordrecht, Holland 1988. p. 178, footnote 31
Last revised on August 18, 2019 at 21:32:24. See the history of this page for a list of all contributions to it.