Are locally contractible spaces hereditarily paracompact?

Question

The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.

I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $\underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $\mathcal F(U) \to \mathcal F^+(U)$ is surjective for all open subsets $U \subseteq X$, assuming the existence part of the glueing axiom.

So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.

Answer 1

A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $\mathbb{R}$, but not paracompact.

Answer 2

The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.

Answer 3

To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility, in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3), where he also explains how Ramanan uses the hereditary paracompactness condition.