Is the pathspace of a CW complex locally compact?

Question

Let $X$ be a CW complex and $I$ denote the closed interval $[0,1]$. Equip the path space $X^I = \operatorname{Map}(I,X)$, the set of continuous map from $I$ to $X$, with the compact-open topology. Is it true that $X^I$ is locally compact? If not, what are the common conditions to make it so? (I ultimately care about the case when $X$ is a manifold.) How about $X^{\mathbb{R}^1}$?

Answer 1

No, $X^I$ is never locally compact unless $X$ is $0$-dimensional (and similarly for $X^{\mathbb{R}}$). If $X$ has positive dimension, then in particular it contains a closed subspace homeomorphic to $I$ and so $X^I$ contains a closed subspace homeomorphic to $I^I$, so it suffices to show that $I^I$ is not locally compact. In this case the compact-open topology is just the topology of uniform convergence. Now take any sequence $(f_n)$ in $I^I$ with no accumulation point (say, $f_n(x)=x^n$). We can scale down this sequence to assume the $f_n$'s are all arbitrarily small in the sup norm, and then translate them to be contained in any open ball in the sup norm. So, $I^I$ is nowhere locally compact: every nonempty open set contains a sequence with no accumulation point.