Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:59:46Z http://mathoverflow.net/feeds/question/6227 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6227/which-topological-spaces-have-the-property-that-their-sheaves-of-continuous-funct Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? Qiaochu Yuan 2009-11-20T06:24:21Z 2009-11-22T21:12:42Z <p>I hope I'm using the terminology correctly. What I mean is this: fix $K = \mathbb{R}$ or $\mathbb{C}$ (I'm interested in both cases). Which topological spaces $X$ have the property that for every open set $U$, every continuous function $f : U \to K$ is a quotient of continuous functions $\frac{g}{h}$ where $g, h : X \to K$ and $h \neq 0$ on $U$?</p> http://mathoverflow.net/questions/6227/which-topological-spaces-have-the-property-that-their-sheaves-of-continuous-funct/6257#6257 Answer by Andrew Stacey for Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? Andrew Stacey 2009-11-20T12:01:38Z 2009-11-22T21:12:42Z <p>This isn't a complete answer, but I think that whatever the family is, it contains compact metric (metrisable) spaces. With a paracompactness argument, I suspect that it would extend to locally compact, and I would not be surprised if one could replace "metrisable" by something weaker (though I think that it would need that separation property one-above-normal which I can never remember the name of: namely that every closed set is the zero set of a continuous function).</p> <p>Here's a proof (I hope): Let $M$ be a compact metric space, $U \subseteq M$ an open subset, $f : U \to \mathbb{R}$ a continuous function. Let's write $K$ for the complement of $U$ in $M$. For each $n \in \mathbb{N}$, let $C_n \subseteq U$ be the subset consisting of points at least distance $1/n$ away from $K$. Then $C_n$ is closed in $M$, hence compact, and $\bigcup C_n = U$. Let $h_0 : M \to \mathbb{R}$ be the "distance from $K$" function (so that $C_n = h_0^{-1}([1/n,\infty))$). Let $V_n$ be the complement of $C_n$.</p> <p>As $C_n$ is compact, $f$ is bounded on $C_n$. Let $a_n = \max\{|f(x)| : x \in C_n\}$, then $(a_n)$ is an increasing sequence. Let $(b_n)$ be a decreasing sequence that goes to $0$ faster than $(a_n)$ increases, specifically that $(a_nb_n) \to 0$. Let $r : [0,\infty) \to [0,\infty)$ be a continuous decreasing function such that $r(1/n) = b_{n+1}$ (as $(b_n) \to 0$ (this always exists) and let $h = r \circ h_0$. Then for $x \in V_{n-1}$, $h_0(x) \lt 1/(n-1)$ so $h(x) \lt b_n$.</p> <p>Then $h : M \to \mathbb{R}$ is a continuous function. Moreover, $h f$ (the product, with $h$ restricted to $U$) has the property that for $x \in C_n \setminus C_{n-1} = V_{n-1} \setminus V_n$,</p> <p>$$|(f h)(x)| = |f(x)| |h(x)| \le a_n b_n$$</p> <p>Thus as $x \to K$, $(f h)(x) \to 0$ and so we can extend $f h$ to a continuous function $g : M \to \mathbb{R}$ by defining it to be $0$ on $K$.</p> <p>Then on $U$, $f = g/h$.</p> <p>(I made this up, so obviously, there may be something I've overlooked in this so please tell me if I'm not correct.)</p> <p><strong>Edit:</strong> This one's been bugging me all weekend. I've even gone so far as to look up <a href="http://en.wikipedia.org/wiki/Perfectly%5Fnormal%5Fspace" rel="nofollow">perfectly normal</a>.</p> <p>This property holds for perfectly normal spaces. In a perfectly normal space, every closed set is the zero set of a function (to $\mathbb{R}$, and this characterises perfectly normal spaces according to Wikipedia).</p> <p>Here's the proof. Let $X$ be a perfectly normal space. Let $U \subseteq X$ be an open set, and $f : U \to \mathbb{R}$ a continuous function. Let $r : X \to \mathbb{R}$ be such that the zero set of $r$ is the complement of $U$. Let $s : \mathbb{R} \to \mathbb{R}$ be the function $s(t) = \min\lbrace 1, |t|^{-1}\rbrace$.</p> <p>The crucial fact is that if $p : U \to \mathbb{R}$ is a <strong>bounded</strong> function then the <em>pointwise</em> product $r \cdot p : U \to \mathbb{R}$ (technically, $p$ should be restricted to $U$ here) extends to a continuous function on $X$ by defining it to be zero on $X \setminus U$.</p> <p>From this, the rest follows easily.</p> <ol> <li><p>The composition $s \circ f$ is bounded on $U$, hence $r \cdot (s \circ f)$ extends to a continuous function on $X$, say $h$.</p></li> <li><p>The product $(s \circ f) \cdot f$ is also bounded on $U$, since $(s \circ f)(x) = \min\lbrace 1, |f(x)|\rbrace)$. Hence $r \cdot (s \circ f) \cdot f$ extends to a continuous function on $X$, say $g$.</p></li> <li><p>As $s(t) \ne 0$ for all $t \in \mathbb{R}$, $(s \circ f)(x) \ne 0$ for all $x \in X$. Hence $h(x) \ne 0$ for all $x \in U$.</p></li> <li><p>Finally, on $U$, $g(x) = h(x) \cdot f(x)$, whence, as $h$ is never zero on $U$, $f = g/h$ as required.</p></li> </ol> <p>This isn't a complete characterisation of these spaces. Essentially, this result holds if there are <strong>enough</strong> continuous functions (as above) on $X$ and if there are too few.</p> <p>As an example of the latter, consider a topological space $X$ where every pair of non-trivial open sets has non-empty intersection. Then there can be no non-constant functions to $\mathbb{R}$, either on $X$ or on any open subset thereof. Hence every continuous function on an open subset of $X$ trivially extends to the whole of $X$.</p> <p>However, there's probably some argument that says that once you have sufficient continuous functions (say, if the space is functionally Hausdorff - i.e. continuous functions to $\mathbb{R}$ separate points) then it would have to be perfectly normal. The difficulty I have with making this into a proof is that there's no requirement that the function $g$ be zero on the complement.</p> <p>Finally, note that metric spaces are perfectly normal so this supersedes my earlier proof. I leave it up, though, in case it's of use to anyone to see the workings as well as the current state. (Actually, for the record I ought to declare that initially I thought that this was <em>false</em> for almost all spaces. However, once I'd examined my counterexample closely, I realised my error and now I'm having difficulty thinking of a <em>reasonable</em> space where it does not hold.)</p> http://mathoverflow.net/questions/6227/which-topological-spaces-have-the-property-that-their-sheaves-of-continuous-funct/6339#6339 Answer by Jose Capco for Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? Jose Capco 2009-11-21T01:21:56Z 2009-11-21T01:43:06Z <p>I tried a bit of thinking, but I haven't worked all the details. I have a hint though that may lead to the answer of your question. You may want to regard the continuous functions over an open set as a ring. This ring is reduced and commutative (thus there is a so-called <em>rational completion</em>) and we could then look at rational completion of them and this may lead to an answer.</p> <p>A good and downloadable reference of this is found <a href="http://at.yorku.ca/i/a/a/l/94.htm" rel="nofollow">here</a>. A classical reference (and also the best one) is the book of Lambek "<a href="http://books.google.com/books?id=LQPid5RSkJsC&amp;dq=Lambek+rings+of+modules&amp;printsec=frontcover&amp;source=bl&amp;ots=KwwoMK3b7w&amp;sig=jQXc%5FedvqN9LQSilXu7kQxfqg20&amp;hl=en&amp;ei=7T8HS6DxApO1sgbT1vCzCg&amp;sa=X&amp;oi=book%5Fresult&amp;ct=result&amp;resnum=3&amp;ved=0CBAQ6AEwAg#v=onepage&amp;q=&amp;f=false" rel="nofollow">Lectures on Rings and Modules</a>" by Lambek (please don't confuse it with the book of Lam, who happens to have the same first 3 letters in his last name, entitled "Lectures on Modules and Rings"), see for instance sections 2.3 and 4.4 of the book.</p> <p>A few years ago, I had written a small entry in Planetmath that characterized <a href="http://planetmath.org/?method=l2h&amp;from=objects&amp;name=RationalExtension&amp;op=getobj" rel="nofollow">rational extensions of commutative reduced rings</a>. And you can use that as an easy definition.</p>