Jason (jcreed) wrote,
Jason
jcreed

-

Aha! I realized in the shower that I can relax the
no-infinite-predecessor-chain condition to finitely
many infinite predecessor chains,
and the injectivity condition -is- required.
Consider:

A B
* * * * *.. * * * *..
| | | | |.. vs. | | | |..
v v v v v.. v v v v..
*->*->*->*->*.. *->*->*->*->*..

I wonder if finite-component u-structures are
Cantor-Bernstein? I'd tend to think maybe so.

Oh, no, of course not. I'm on crack.
A B
->
* * *->*
<-

-> ->
* * vs. * *
<- <-

-> ->
* * * *
<- <-
.... ....

Okay, so also: suppose f is a u-homom from U to V.
if f isn't mono, then obviously
it isn't injective. If it isn't injective, say
f maps a and b to the same thing, then we can use
a copy of *N and map 0 by x to a and by x' to b,
and the mappings' behavior on the rest of *N is forced.
Moreover, f o x = f o x' by induction. So in any
full subcategory of uStruct that contains *N, the monos
are exactly the injective homomorphisms.
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