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.