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reviewed  Approve Analogue to covering space for higher homotopy groups? 
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revised 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
Or actually, more simply: (7) automatically holds for the endpoints of the interval, and it holds for all other values of $a$ and $b$ by continuity and (5) (since $M(a,b)a$ and $M(a,b)b$ must have constant sign if we stay away from the diagonal). I guess this argument is slightly weaker than the one in my previous comment because it also requires (2). But assuming (2), it shows that (7) also follows from (5) and (8) in an open interval as long as you know that (7) holds for any single $a$ and $b$. 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
Oh, I see, that makes it a lot trickier than I thought. Incidentally, I think (7) follows from (5) and (8), at least on a closed interval. For instance, if $a<b<M(a,b)$, then by continuity and (5) $M(a,c)>c$ for all $c>a$. But now we get a contradiction by taking the limit of iterating $c\mapsto M(a,c)$. 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
Your 5element algebra fails to satisfy (3): $M(e,M(b,c))\neq M(M(e,b),M(e,c))$. 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
I'm pretty sure (3) doesn't imply (4) in the free algebra on 3 generators: there seems to be no way to transform $M(M(M(a,a),M(a,b)),M(M(b,c),M(c,c)))$ that changes the composition of the two halves. 
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revised 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
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answered  Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent? 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
Note that besides the arithmetic and geometric means satisfying (6), there are also the means $((x^p+y^p)/2)^{1/p}$ for any $p\neq0$; the geometric mean can be considered as the limit of these as $p\to 0$. 
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comment 
Cantor's theorem for presheaves?
@DavidRoberts: That assumption is already implicit in the question itself. The question is about categories living inside a single universe, and so $\mathbf{Set}$ and $\mathbf{C}$ are both classes in that universe. We only go outside the universe in order to form functor categories between them. 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
@YaakovBaruch: That fails associativity (weak associativity, even). Consider $M(M(M(a,b),b),M(M(a,b),a))$. 
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comment 
Are all well behaved “mean” functions on $\mathbb{R}^+$ equivalent?
Isn't Q4 pretty much immediate by starting with two arbitrary points and iterating $M$ on them and then comparing that with the dyadic rationals? 
Dec 22 
comment 
Cantor's theorem for presheaves?
In retrospect, I'm not so sure that retractions are important to the argument anymore. Now that I've come up with the new argument, it's clear that the "exhaustions" I was using were just shadows of a much more natural structure: an unbounded functor $G:\mathbf{C}\to\mathbf{Set}$ (the exhaustion then being $F(A)=G(A)$). If you forget the functorial structure of $G$ and remember only $F$, you need to regain functoriality using representable functors, which is why my original argument required local smallness. 
Dec 22 
revised 
Cantor's theorem for presheaves?
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Dec 22 
revised 
Cantor's theorem for presheaves?
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Dec 22 
revised 
Cantor's theorem for presheaves?
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Dec 22 
answered  Anticompactness 