Re: are fibrations evil?

[John Baez <baez@math.ucr.edu>]

Thomas wrote, of fibrations:

Sure, they are "evil" but it seems to be beneficial to be "evil" sometimes.
>

Of course!  Anybody who thinks that concepts should be avoided merely
because they're "evil" in the technical sense must also think that odd
numbers are peculiar and perfect groups are the best groups to study.  I'm
finding it quite amusing how people are getting worked up over the concept
just because its name has moral overtones.

An excellent example of an evil but useful concept is the concept of
"skeletal category".  Every category is equivalent to a skeletal one
(assuming choice), but not every category is skeletal.  So, this concept is
evil.  But it's sometimes nice to prove a theorem for all categories by
proving it for skeletal categories.

We can formalize this a bit.

Remember, a property P of objects in an n-category is "evil" if there's an
object with that property that is equivalent to an object without that
property.  It's "non-evil" if whenever x has property P and x is equivalent
to y, y also has that property.

(I'm using classical logic here.  If we're using intuitionistic logic, the
really useful concept is the concept that I'm calling "non-evil".  We should
probably call it "good": evil is just the absence of good.  But "non-evil"
is less likely to cause confusion.)

Every property P can be made non-evil by defining a new property P' that
means "equivalent to something with property P".  Sometimes it's more
convenient to work with objects with property P than objects with property
P'.  And note: if the property Q is non-evil, the theorem

all objects with property P have property Q

automatically implies

all objects with property P' have property Q

So, for example, if Q is a non-evil property of categories, to prove Q holds
for all categories it suffices to prove it for skeletal categories.

Or - with a bit more work to fill in the details - if Q is a non-evil
property of functors, to prove Q holds for all Street fibrations it suffices
to prove it for fibrations.

Best,
jb

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