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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 $B$.

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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 $\chi(S^3) = 0$.

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    $\begingroup$ If you don't like that both Euler characteristics were $0$ here you can add a point to both $S^1$ and $S^3$ but I still don't think you get a reasonable number. Everything would work out much more nicely if the Euler characteristic of $\Omega X$ were the inverse of the Euler characteristic of $X$, but of course this is very far from true. Loosely speaking the problem is that the Euler characteristic is "additive" (behaves nicely wrt some homotopy colimits) but taking mapping spaces is "multiplicative" (behaves nicely wrt homotopy limits). $\endgroup$ Oct 4, 2015 at 16:05

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