Hi,

I'm working through MacLane's "Categories for the Working
Mathematician" (although as I'm still an unemployed student maybe I'm
not in the target audience!) Anyway, in a set of exercises about mono
and epi arrows, he asks you to prove that in the category of sets,
every idempotent splits.

Now I can't work out how one might go about proving this, and am even
not sure if I've misunderstood the question because of the following
example. Take 2 = {0,1} a set with two elements and let f: 2 -> 2
where f(x) = 0. Clearly f is idempotent since f(f(x))=f(x)=0 but you
can list all the (4) maps from 2->2 and it doesn't seem that there are
any such that f=gh and hg=Id, which would seem to be implied by what I
thought I was trying to prove!

Can anyone see what I've misunderstood? I'm a bit stuck!

Rupert

On Apr 29, 7:13 pm, Rupert Swarbrick <rswarbr...@gmail.com> wrote:
Hi,

I'm working through MacLane's "Categories for the Working
Mathematician" (although as I'm still an unemployed student maybe I'm
not in the target audience!) Anyway, in a set of exercises about mono
and epi arrows, he asks you to prove that in the category of sets,
every idempotent splits.

Now I can't work out how one might go about proving this, and am even
not sure if I've misunderstood the question because of the following
example. Take 2 = {0,1} a set with two elements and let f: 2 -> 2
where f(x) = 0. Clearly f is idempotent since f(f(x))=f(x)=0 but you
can list all the (4) maps from 2->2 and it doesn't seem that there are
any such that f=gh and hg=Id, which would seem to be implied by what I
thought I was trying to prove!

Can anyone see what I've misunderstood? I'm a bit stuck!

Rupert

Notice that if f : X -> X is idempotent,
a splitting for f is a pair of arrows

(h : X --> Y, g : Y --> X)

such that g h = f and h g = id_Y.
You seem to be looking for a splitting
with Y = X?

In your specific example, you can take
Y = { 0 }, h : X --> Y the constant
map and g : Y --> X the inclusion.

-- m