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This is a question about a proof in the article Group Cohomology with Lipschitz Control and Higher Signatures by Connes, Gromov and Moscovici, which may be found here. Namely, their "metric extension lemma" asserts that one can extend a metric from a subspace of a product to the whole space.

The precise statement is as follows and may be found on page 41 of the article:

Let $Y$ and $P$ be locally compact metrizable spaces, and let $Z\subset Y\times P$ be closed. Let $d$ be a metric on $Z$ such that $d|Y\times p=d_Y$ for all $p\in P$ and a given metric $d_Y$ on $Y=Y\times p$. Then there exists a metric $\bar d$ on $Y\times P$ such that (i) $\bar d|Y\times p=d_Y$ for all $p\in P$ and (ii) $\bar d|Z\geq d$.

I added the condition that $P$ is metrizable and that $Z$ is closed to the statement of the theorem, and both conditions are obviously necessary (Otherwise, one could, for instance, take $Z=(0,1)\subset[0,1]=P$ and $Y=\{*\}$, and the lemma would imply that every metric on $(0,1)$ can be bounded above by a metric on $[0,1]$.)

In the article, the authors first prove the statement in the case that $Y$ and $P$ are compact. This proof seems to be OK to me.

However, when they want to reduce the general case to the compact one, they take a locally finite cover of $Y\times P$ by compact product subsets $Y_i\times P_i$ (as YCor pointed out, every metric space is paracompact.) and equip each of them with a metric as constructed in the compact case. Then they define the final metric as the supremum of the metrics which are bounded above by the constructed metric on every $Y_i\times P_i$, and which equal $d_Y$ on every slice $Y\times p$.

I have three questions regarding this proof:

  1. I think one could have arbitrary metrics on $Y_i\times P_i$ if this product is small enough to contain only one point of $Z$. So how can the result have anything to do with $Z$?
  2. Can one explicitly describe this "supremum metric"? To be precise, how does one know that the set of metrics with the given properties is non-empty?
  3. Can we still prove the lemma in the general case, perhaps adding some condition on $Z$, say, $Z\subset Y\times p$ being coarsely dense for every $p\in P$?
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  • $\begingroup$ There are several definitional issues I'd first need to ask. (1) for $X$ a topological space, does "a metric on $X$" precisely mean a continuous function on $X\times X$ satisfying the axioms of a distance? "$d$ a metric on $Z$ such that $d_{Y\times p}=d_Y$ for every $p$": this seems senseless, should it be understood as " $d_{(Y\times p)\cap Z}=d_Y$ for every $p$"? $\endgroup$ – YCor Nov 13 '17 at 19:02
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    $\begingroup$ Stone: every metrizable space is paracompact. $\endgroup$ – YCor Nov 13 '17 at 19:05
  • $\begingroup$ "So how can the result have anything to do with $Z$"? I guess this is precisely hidden in this awkward wording "and which equal $d_Y$ on every slice $Y\times p$". It should be "and which equal $d_Y$ on every slice $(Y\times p)\cap Z$". $\endgroup$ – YCor Nov 13 '17 at 19:10
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    $\begingroup$ "how does one know that the set of metrics with the given properties is non-empty". If Gromov wrote this, it's likely that by metric he does not care with the inconvenient separability axiom. So the non-emptyness is obvious: just consider $\delta((y,p),(y',p'))=d_Y(y,y')$. $\endgroup$ – YCor Nov 13 '17 at 19:17
  • $\begingroup$ By "a metric of $X$", I suppose they mean a metric which induces the given topology. However, a continuous function satisfying the axioms of a metric (possibly without separability) would be enough for the application in their paper. $\endgroup$ – Benedikt Hunger Nov 14 '17 at 6:13

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