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Below is an image of Hatcher Proposition 4G.2, which is used to prove the Nerve Theorem (Hatcher Corollary 4G.3).

enter image description here

My question is about the sentence that is highlighted in yellow. I thought that the existence of a partition of unity subordinate to any open cover of a space requires paracompactness and the Hausdorff property, not just paracompactness alone.

For example, there exist paracompact spaces that are non-Hausdorff and don't necessarily admit partitions of unity. See, e.g., this MO post.

Question: In order for the highlighted sentence to be correct, do we need the assumption that $X$ is Hausdorff? If so, is the Proposition false for paracompact non-Hausdorff spaces? If not, what I am missing?

Further research: In an early (perhaps earliest) formulation of the nerve theorem for good open covers, due to Borsuk, it is assumed that $X$ is a metric space, hence paracompact and Hausdorff. In another formulation, due to Weil p.141, it is assumed that $X\times X \times [0,1]$ is normal. In a later citation by McCord (Theorem 2), it is noted that this assumption can be replaced with one that $X$ is a "separable metric." However, in none of these references is it assumed that $X$ is simply paracompact. So far, the only reference I've found with this assumption alone is Hatcher (and many others that cite Hatcher). But I'm not convinced the partition of unity argument works unless $X$ is Hausdorff.

R. H. Vellstra
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    I don't know what definition of paracompact Hatcher is using, but some people (very rare though) assume Hausdorfness to be part of the definition.

    Barring that, the partition of unity argument does need Hausdorf since being paracompact and Hausdorff is equivalent to every open cover admiting a subordinate partition of unity.

     – HackR Nov 14 '24 at 21:33
  • @HackR I just discovered this as well-- I referenced Bredon's Topology and Geometry in which the definition of paracompactness is given only for Hausdorff spaces. Hatcher never gives a definition of paracompact, but I guess he must've had that one in mind. – R. H. Vellstra Nov 14 '24 at 21:35
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    @HackR This is not rare. For example, Engelking and Dugundji do it. Munkres writes " Many authors, following the lead of Bourbaki, include as part of the definition of the term paracompact the requirement that the space be Hausdorff. (Bourbaki also includes the Hausdorff condition as part of the definition of the term compact.) We shall not follow this convention." – Paul Frost Nov 14 '24 at 23:48
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    One can definitely say that Hatcher's proof requires the existence of partitions of unity. This follows from paracompact Hausdorff, but certainly there are spaces with partitions of unity which are not paracompact Hausdorff. – Paul Frost Nov 14 '24 at 23:54
  • By the way, each $T_1$-space which has partitions of unity is paracompact Hausdorff. For a proof see e.g. Engelking "General Topology" Theorem 5.1.9. – Paul Frost Nov 15 '24 at 10:13

1 Answers1

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It is not uncommon to use the term 'paracompact' to refer exclusively to Hausdorff spaces, in which every open cover has a locally finite refinement. For Hausdorff spaces, this condition is equivalent to every open cover admitting a subordinate partition of unity. That this was Hatcher's intent can be confirmed by looking at a different book of his, Vector Bundles and K-Theory, where he defines paracompactness by the latter condition on p.35.

Thorgott
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  • In his AT-book Hatcher often refers to his other book [VBKT] = Vector Bundles and K–Theory. This supports your answer. – Paul Frost Nov 14 '24 at 23:44