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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.

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.