(This question was originally asked on StackExchange: http://math.stackexchange.com/questions/82437/can-one-average-close-smooth-functions )

Suppose $M$ is a connected, smooth, second-countable manifold.

Let $U \subset M^n$ be some neighbourhood of the diagonal. We will call a function $a: U \times \Delta_n \rightarrow M$ an "averaging operator of order n" if $a|_{U \times v_i} = \pi_i$, where $v_i$ denotes the i-th vertex of the standard $n$-simplex $\Delta_n$.

Intuitively, $a$ should tell us how to take a weighted average of "close" functions.

Does every $M$ admit a smooth averaging operator of order $n$, for some $U$ and all $n$?

(I am trying to prove that smooth functions are dense in $Hom(M,N)$ without embedding $N$ into any $\mathbb{R}^n$ and I think it would easily follow from the above.)