# Why should have Peter May worked with CGWH instead of CGH in “The Geometry of Iterated Loop Space”?

This is a follow-up to Dan Ramras' answer of this question.

The following correction can be found in the errata to The Geometry of Iterated Loop space (Page 484 here).

The weak Hausdorff rather than the Hausdorff property should be required of spaces [...] in order to validate some of the limit arguments used [...].

I was not able to figure out which type of "limit arguments" really would have needed the weak Hausdorff condition instead of the regular one used in the original paper and would be happy to understand the necessity of the correction, ideally by means of an explicit example in the paper.

• Maybe Paragraph around prop 2.4.22 of Hovey's model category book helps a little here. It explains construction of limit and colimit in CGWH and why it is a better category. I am wondering if we work in CGH, the model structure may not be equivalent to the one on Top, but I do not know. – Mingcong Zeng Apr 28 '15 at 17:43
• Not an answer, but one should mention the obvious -- philosophically, one should expect CGWH to work better because the WH condition (the diagonal $X \to X \times X$ is closed when the product is taken in CG spaces) is stated entirely in terms of the CG category, whereas the H condition (the diagonal is closed when the product is taken in Top) refers us back to Top. So there's a "mismatch" in the definition of CGH -- it's like defining a scheme to be separated if its underlying space is Hausdorff, which is totally wrong. Plus, we're potentially re-importing pathologies from Top... – Tim Campion Oct 20 '15 at 17:35

I don't know for Peter May's work. I know that the weak Hausdorff condition can be useful for the following reason: let $f:X\to Y$ be a continuous map between Hausdorff k-spaces. Then consider the equivalence relation $\mathcal{R}_f$ on $X$ associated with $f$: $x \mathcal{R} y$ if and only if $f(x)=f(y)$. The graph of $\mathcal{R}_f$ is closed in $X\times X$ since it is the inverse image by $(f,f)$ of the diagonal of $Y$ which is closed in $Y\times Y$ because $Y$ is Hausdorff. The quotient set $X/\mathcal{R}_f$ equipped with the final topology is weak Hausdorff. There are examples (that I don't have in mind, maybe someone will be able to refresh my memory) where the quotient set $X/\mathcal{R}_f$ equipped with the final topology is not Hausdorff anymore. Of course, the "Hausdorffisation" of $X/\mathcal{R}_f$ still exists but the underlying set is not the quotient set anymore. More identifications can be done by the "Hausdorffisation" functor, i.e. we don't have the control anymore over what is identified in $X/\mathcal{R}_f$.