[Haskell-cafe] morphisms in IO

[Dan Weston]

I truly have no idea what you are saying (and probably not even what I 
am saying), but I suspect:

a) You are calling IO the target category of applying the functor IO 
[taking a to IO a and (a->b) to (IO a -> IO b)] to Hask.

b) This category is hardly bereft, nor discrete. Its morphisms are IO a 
-> IO b. Maybe you are thinking that it becomes an empty category after 
some fixed point process where you strip off the IO with repeated joins 
(or runIOs)? Remember there can be arbitrarily many nestings of IO, so 
that main :: IO (IO (IO Int)) is (a program that when run returns)^3 an 
Int. This is a stream with no finite least fixed point.

c) What you are calling a "bereft" category is an empty category. 
Without (identity) morphisms, there can be no objects. There is only one 
such category (the empty category), so naturally any two such are 
isomorphic (for what it's worth, which I suspect is not much).

Dan

Gregg Reynolds wrote:
> I'm working on a radically different way of looking at IO.  Before I
> post it and make a fool of myself, I'd appreciate a reality check on
> the following points:
> 
> a)  Can IO be thought of as a category?  I think the answer is yes.
> 
> b)  If it is a category, what are its morphisms?  I think the answer
> is: it has no morphisms.  The morphisms available are natural
> transformations or functors, and thus not /in/ the category.
> Alternatively: we have no means of directly naming its values, so the
> only way we can operate on its values is to use morphisms from the
> outside (operating on construction expressions qua morphisms.)
> 
> c)  All categories with no morphisms ("bereft categories"?) are
> isomorphic (to each other).  I think yes.
> 
> Thanks,
> 
> gregg
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