My friend posed a question that how we can generalize the notion of mean value on topological spaces, preferably Poincare complexes, $X$, so that it satisfies cheap properties of mean value on Euclidean spaces. More precisely, we want sequence of continuous maps $ m_n:X^n\rightarrow X$ for every $n$ such that it is symmetric ( or if you like $m_n: SP_n(X)\rightarrow X$ ) and also $ m_n\circ \Delta = id_X$ meaning that mean value of identical points should be the point itself. It seems even these properties are highly restrictive on topology, for example it is easy to show that we cannot define mean value on spheres. I wonder if there exists nonacyclic finite dimensional complex that admits mean value function?
I am not sure whether "acyclic" means connected for you. If yes, then there are obvious counterexamples (finite spaces with the discrete topology). So, assume that the space is connected. If you assume that the space $X$ is a compact, connected polyhedron, then the existence of even one $m_n,$ for $n\geq 2,$ then $X$ is contractible. This is a Theorem of Eckmann (1954). Both assumptions are necessary: The EilenbergMcLane Space $K(\mathbb{Q}, m)$ admits an $n$mean for EVERY $n$ (so is an example of the sort you seek), but is not compact. The $n$solenoid admits an $n$mean but is not a polyhedron. For, much, much more on this subject see: MR1490716 (98k:55013) Hilton, Peter(1CFL) A new look at means on topological spaces. (English summary) Internat. J. Math. Math. Sci. 20 (1997), no. 4, 617–620. 


I think there's a mod two homology obstruction. Let's assume $X$ is a $k$dimensional manifold, compact and without boundary, and let $\mu \in H_k(X, \mathbb Z_2)$ be its fundamental class. Then $\mu$ maps to some element of $H_k(X^n,\mathbb Z_2)$ under $\Delta_*$, and by your symmetry assumption, it has to lie in the diagonal subspace of $H_k(X^n,\mathbb Z_2) \simeq \prod_n H_k(M;\mathbb Z_2)$ (Kunneth), which means that when you further apply $m_{n*}$, the element of $H_k(X,\mathbb Z_2) \simeq \mathbb Z_2$ you get would be represented by a multiple of the integer $n$. So provided $n$ is even, this is a contradiction. If your manifold were orientable it would give you an obstruction for any $n \geq 2$. 

