I will have another go at arguing that dagger-categories are *not* evil.

Let’s look at a simpler case first. Consider the property “$1 \in X$” on sets. As a property of abstract sets, this is evil: it’s not invariant under isomorphism, e.g. any iso $\{1,2\} \cong \{2,3\}$.

But it is manifestly non-evil as a property of, say, “sets equipped with an injection to $\mathbb{N}$”. Looking at this gives a non-evil structure on abstract sets: “an injection $i : X \to \mathbb{N}$, such that $1 \in \mathrm{im}\ i$”.

Ah (you may say) so if this is non-evil, it can’t reflect our original idea correctly: it would have to transfer along that $\{1,2\} \cong \{2,3\}$. But that’s not such a clear-cut complaint. As a structure on abstract sets, it does transfer along that isomorphism. But we were thinking of them from the start not just as abstract sets, but as subsets of $\mathbb{N}$, i.e. as *already* equipped implicitly with injections to $\mathbb{N}$. Considered as such, one of them contains $1$ and the other doesn’t; and there’s no isomorphism between them which commutes with those injections.

Summing up: “containing 1” is certainly non-evil as a property of “sets with a mono to $\mathbb{N}$”. This induces a non-evil property/structure on abstract sets which you may or may not agree matches our original idea, because it requires considering different monos to $\mathbb{N}$ besides the ones we were already (implicitly) thinking of.

Now, back to dagger-categories. It seems reasonable that dagger-categories are evil *when regarded as structure on categories*. The post by Peter Selinger linked by Simon Henry argues this quite persuasively: proving a specific no-go theorem, showing there can exist no notion of dagger-structure satisfying certain desirable properties.

However, *structure on categories* is not the only way to look at dagger-categories. They can instead be seen as structure on “pairs of categories connected by a faithful and essentially surjective functor $i : \mathbf{C}_u \to \mathbf{C}$”. (Full definition below.)

You may be thinking: this can’t be right, because it induces a non-evil structure on categories (“equip with a fully faithful inclusion from some other category, and then the dagger-structure”) which would violate Selinger’s no-go argument. However, it doesn’t: this structure doesn’t allow a definition of unitary maps in the sense Selinger’s argument assumes. Given $A, B \in \mathbf{C}_u$, we can say a map $iA \to iB$ is unitary if it’s the image of some map $A \to B$. But given just $A, B \in \mathbf{C}$, this unitariness isn’t well-defined for $f : A \to B$; different ways of expressing $A$ as $iA'$ and $B$ as $iB'$ might give different answers as to whether $f$ is unitary.

Expressed in this form, transferring the “weak dagger structure” on $\mathbf{fdHilb}$ along the equivalence to $\mathbf{fdVect}$ yields a weak dagger structure where the functor $i$ is not injective on objects. Selinger’s argument shows that something like this is unavoidable.

Summing up again: dagger-structure is certainly not evil when viewed as a structure on “categories with a distinguished faithful, ess. surj. inclusion”. This gives a definition of weak dagger structure as a non-evil structure on categories, which you may or may not accept, because it requires us to loosen up our original expectation that the “subcategory” of unitary maps should be literally bijective on objects, i.e. to go beyond the kind of “subcategories” we were originally thinking of.

Full definition: a *weak dagger category* may be taken to consist of:

- a category $\newcommand{\C}{\mathbf{C}}\C$;
- a groupoid $\C_u$, with a faithful and essentially surjective functor $i : \C_u \to \C$;
- a functor $\dagger : \newcommand{\op}{\mathrm{op}}\C^\op \to \C$;
- a natural isomorphism $\varphi : \dagger \cdot i^\op \cong i \cdot (-)^{-1} : \C_u^\op \to \C$;
- a natural isomorphism $\psi : \dagger \cdot \dagger^\op \cong 1_\C$;
- such that for all $A \in \C_u$, $\psi_{iA} = \varphi_A (\varphi_{A}^\dagger)^{-1} : (iA)^{\dagger \dagger} \to iA$, or equivalently, as natural transformations, $\psi \cdot i = (\varphi \cdot (-)^{-1})(\dagger \cdot \varphi^\op)^{-1} : \dagger \cdot \dagger^\op \cdot i \to i$;
- and such that for any $A, B \in \C_u$, the image of $\C_u(A,B) \to \C(iA,iB)$ consists of
*all* $u$ such that $\varphi_A \cdot u^\dagger \cdot \varphi_B^{-1}$ is a 2-sided inverse for $u$.

Given this, I claim:

- each component is “non-evil” as structure on the earlier components (this can be made precise as a lifting property for forgetful functors between 2-categories);
- strict dagger categories are precisely weak dagger categories such that $i$ is bijective on objects and $\varphi_A = 1_{iA}$ for all $A \in \C_u$;
- given any weak dagger cat with $i$ bijective on objects, one can modify $\dagger$, $\varphi$, and $\psi$ to obtain a strict dagger-structure, equivalent to the original, with the equivalence acting trivially on $\C$, $\C_u$, and $i$;
- therefore, given a category $\C$ equipped with a distinguished all-objects subgroupoid $\C_u$, “strict dagger structures on $\C$ with unitaries $\C_u$” correspond to “weak dagger structures on $\C_u \to \C$”; so strict dagger structure is non-evil as a structure on “cats with a distinguished all-objects subgroupoid”;
- if we drop the last component of the definition, we similarly get a non-evil structure of “strict dagger structures on $\C$ with unitaries including at least $\C_u$”;
- given any weak dagger cat $\C$, there is an equivalent strict one, with the same unitary category $\C_u$, but with new ambient category given by the objects of $\C_u$ with the morphisms of $\C$. So overall, weak dagger categories do not give us anything essentially different from strict ones.

(I’m pretty sure that I’m remembering most of the ideas here from somewhere, but I can’t find where. The nearest I can find is this post by Mike Shulman on that same categories list thread. Better references very welcome.)

notcoincide with the "maps which make two objects the same". This seems to be the core of what is going on in the link in Simon Henry's last comment. $\endgroup$ – Eric Wofsey Oct 5 '15 at 16:0111more comments