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1 point by rntz 3414 days ago | link | parent

rev isn't idempotent, however, rev:rev (rev composed with itself) is, which is I suspect what fallintothis was thinking. "Involutary" (which term I've never run across prior to this) appears to be a term for just this property: that the square is idempotent.


2 points by shader 3414 days ago | link

Involution is more strict than the square being idempotent; it also must be the identity function. I.e. abs^2 is idempotent but abs itself is not involutary, since abs^2(-1) = 1 != -1

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