What is the Euler characteristic of a mapping space?

Question

Suppose that $A$ and $B$ are topological spaces homotopy equivalent to finite cell complexes, and let $B^A = \mathrm{maps}(A,B)$ denote the space of maps from $A$ to $B$. Is it there a formula for the Euler characteristic $\chi(B^A)$ of this mapping space in terms of, say, the Euler characteristics of $A$ and $B$ (or similar not-too-difficult data, like the Betti numbers)? I think that if $\chi(B^A)$ were $\chi(B)^{\chi(A)}$, then Wikipedia would have said so, and I would have known about it.

I care most about the case when $B$ is a manifold and $A$ is a surface, which is fairly specific, so I will accept answers that demand good behavior of $A$ and/or $B$.

Answer 1

This Euler characteristic usually won't be well-defined. For example, take $A = S^1$ and $B = S^3$. Then the mapping space $[A, B]$ is the free loop space $L S^3$, which decomposes as a product

$$L S^3 \cong S^3 \times \Omega S^3$$

because $S^3$ has a Lie group structure. This makes the rational cohomology of $L S^3$ easy to compute: it's the tensor product of the rational cohomology of $S^3$ and the rational cohomology of $\Omega S^3$, and the latter is free on a generator of degree $2$. Hence the Poincare series of $L S^3$ is

$$\frac{1 + t^3}{1 - t^2}$$

and in particular $LS^3$ has nonvanishing rational cohomology in arbitrarily high degrees (which is usually the case with free loop spaces). Removing the common factor of $1 + t$ and plugging in $t = -1$ for kicks gets us $\frac{3}{2}$, which doesn't have any obvious relationship to $\chi(S^1) = 0$ or $\chi(S^3) = 0$.