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From the proof of 1 it is clear that order preservation doesn't require preservation of bottom. A similar proof uses preservation of bottom instead of top, but if both conditions are dropped the map needn't even be monotonic. Let I = [-1,1] and let C(I) -> I be

Let I be a closed interval and let C(X) denote the bipointed midpoint algebra of continuous functions from a space X to I. I would not have guessed the following: 1. Any map from C(X) to C(Y) that preserves midpoints, top, and bottom is monotonic and preserves constant functions. The order preservation is an immediate consequence of: f =< g iff there exists h such that f|top = g|h If we specialize to the case that X and Y are single points, the order preservation implies the only map from I to I that preserves midpoints, top and bottom, is the identity map (after first noticing that it must have a dense subset of fixed points corresponding to the dyadic rationals). Let I -> C(X) be the "K-map" that assigns constant maps. For any y:Y let C(Y) -> I be its evaluation map. Then for any map C(X) -> C(Y) that preserves midpoints, top, and bottom we have that I -> C(X) -> C(Y) -> I is the identity map. And from that we can easily conclude that C(X) -> C(Y) preserves constant maps. In a 22 Dec posting I said that the "mean-value" function M:C(I) -> I can be characterized from its order preservation together with what it does to constant functions and MxM C(I v I) --> C(I + I) --> C(I) x C(I) ---> I x I C(F) | | m v v M C(I) -----------------------------------> I where F:I -> I v I is the coalgebra structure and m is the midpoint operation. I went on to say that if C(F) is inverted then this diagram can be read as a fixed-point definition of M: "It's the unique fixed-point of an operator acting on the set of all those order-preserving maps from C(I) to R that do the right thing on constant functions." It's much better to change the R to I and say just that it's acting on the set of bipointed midpoint-algebra homomorphisms. To recapitulate (using \ for lambda, T for top, B for bottom): 2. The mean-value operator, M:C(I) --> I, is the unique map such that: M(\x.B) = B M(\x.T) = T (Mf)|(Mg) = M(\x.(fx)|(gx)) Mf = (M(\x.f(B|x)))|(M(\x.f(x|T))) No inequalities, no limits, only equations. Let me take the occasion to do a little clean-up. In a 31 Dec posting I gave a curse-of-analysis proof for: 3. Midpoint-preserving functions between intervals are monotonic. Here's a much better proof. Monotonicity is equivalent to the preservation of betweenness. Suppose f is a midpoint-preserving map between intervals. It's conceptually useful to take the target to be, instead, the real line and show that a failure to preserve betweenness forces the values of f to be unbounded. So let a,b,c be points in the source interval with b between a and c which fact will be denoted as {a.b.c} and suppose that fb is not between fa and fc. Without loss of generality we can assume that {fa.fc.fb}. There exists {a.b'.c} such that either b = a|b' or b = c|b'. Each of the equations fb = fa|fb' and fb = fc|fb' implies {fa.fc.fb'}. So {a.b'.c} is still an example but the first equation implies that the distance from fc to fb' is twice -- and the second equation implies it is _at least_ twice -- the distance from fc to fb. By iteration we may thus increase that distance beyond any bound. A point that the previous proof didn't really cover is dispatched by this corollary: 4. Midpoint-preserving functions between intervals are either one-to-one or constant. Suppose a and c are distinct but fa = fc. If there's c' such that c = a|c' then we may infer that fa = fc'. By replacing c with c' in this manner as long as possible we may assume that a and c are such that there is no solution to the equation c = a|x and from that we may infer that if c' is chosen to be the endpoint such that {a.c.c'} then there is {a,b,c} such that c = b|c'. By the last result we have fa = fb = fc hence fc = fb = fc'. Thus we may assume that a and c are distinct with fa = fc and c an endpoint. Similarly we may assume that a is an endpoint and finish by applying, once again, the last result. Besides injectivity we have this corollary on surjectivity obtained by combining 1 and 3: 5. The image of any midpoint preserving function between intervals is a closed interval. PS the map that sends f to f(-1)|(-f1). It sends all constant functions to 0, it sends the identity map to -1 and it sends the negating map to 1.) The proof of 3 suggests that if the target interval is replaced by the reals then there do exist midpoint-preserving maps that are not monotonic. And that is, indeed, the case in the presence of the axiom of choice: view the reals as a rational vector-space and count the number of endomorphisms (each of which preserves midpoints) to find that there are 2^{2^N} of them. But there are only 2^N monotonic self maps (midpoint preserving or not). Do we need the axiom of choice? If, instead, we add the axiom of measurability, then the counterexamples disappear: let f be a measurable midpoint-preserving map from R to R; we can easily specialize to the case that f0 = 0, hence f is a measurable group endomorphism; for any real a consider the induced group homomorphism from R/aZ to R/(fa)Z; any measurable group character is continuous and that's enough to force f to be continuous.