true. but the second (third, fourth,nth) hash does not have fewer possible outcomes than the first.

Exactly.

Exactly.

Exactly.

Something else to consider: part of the reason that hashes are non-de-hash-able (or, like, whatever) is that they're lossy. Flip this bit, rotate these ones, log() that one,throw away every fifth value that is a power of two. (Random example, of course, not any real algorithm.)