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Let $\mathbf{Top}$ be the category of topological spaces, and let $I\in\mathbf{Top}$ be the unit interval $I=[0,1]\subset\mathbb{R}$. For any space $X$, let $|X|$ denote the underlying set of points; similarly if $f\colon X\to Y$ is continuous, $|f|$ denotes the underlying function. Define a curve in $X$ to be a continuous map $I\to X$.

Given spaces $X$ and $Y$ and a function $f\colon |X|\to |Y|$ on underlying sets, say that $f$ sends curves to curves if composing $f$ with any curve $c\colon I\to X$ yields a (continuous) curve in $Y$. Of course, if $f$ is continuous then it sends curves in $X$ to curves in $Y$, but the converse may not hold.

I'll say that $X$ has enough curves if, for any $Y$ and function $f\colon|X|\to|Y|$, we have that $f$ is continuous if and only if it sends curves to curves.

An example of a space that does not have enough curves is the "sequence space" $S=\{0\}\cup\{\frac{1}{n+1}\mid n\in\mathbb{N}\}\subset I$. All curves in $S$ are constant, so they cannot detect non-continuous maps out of $S$.

Question: Which well-known classes of spaces have enough curves?

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    $\begingroup$ If I understand this correctly, this is the same as saying that $X$ has final topology w.r.t. the family of maps $C_X=\{ c\colon I\to X; c\text{ is continuous}\}$. I.e., $f\colon X\to Y$ is continuous if and only if $f\circ c$ is continuous for every $c\in C_X$. This also implies that $X$ is a quotient of disjoint sum of several copies of $I$ (one for each $c\in C_X$), therefore $X$ must be a sequential space. $\endgroup$ – Martin Sleziak Jun 16 '16 at 5:01
  • $\begingroup$ I am probably missing something, but it seems if we take $Y$ discrete then any function $f \colon X\to Y$ maps curves to curves (since any $c\colon I\to Y$ is continuous). Doesn't this mean that any space with enough curves must be discrete? (We take $Y$ to be the discrete space on the set $X$ and $f=id_X$.) $\endgroup$ – Martin Sleziak Jun 16 '16 at 5:14
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    $\begingroup$ @MartinSleziak not necessarily. According to Theorem 4.11 (5), pp. 39-40 of the book of A. Kriegl and P.W. Michor, The Convenient Setting of Global Analysis (AMS, 1997), $X$ has the final topology with respect to continuous curves if $X$ is the strong dual of a Fréchet-Montel locally convex vector space. Such spaces are far from being discrete - take e.g. $X=\mathbb{R}^n$ or $X=$ the space of distributions of compact support on a second countable smooth manifold. $\endgroup$ – Pedro Lauridsen Ribeiro Jun 16 '16 at 5:51
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    $\begingroup$ @PedroLauridsenRibeiro I see. It is quite an embarrassing mistake - if I wanted all functions to preserve curves, I had to take $Y$ indiscrete rather than discrete. Of course, the rest of the argument falls apart. So my second comment is completely irrelevant. (But I should probably still let it there, so that other users see what you are reacting to.) BTW the part you found in Kriegl's and Michor's book seem to be rather relevant for this question, at least if we restrict to the case that $X$ is locally convex topological vector space. $\endgroup$ – Martin Sleziak Jun 16 '16 at 7:36
  • $\begingroup$ @MartinSleziak I think that your first comment and js21's answer below (which is partly based on the former) go in a similar direction as Theorem 4.11 of Kriegl-Michor - after all, topological vector spaces are always locally path connected and the example I gave is sequential (as it should). Moreover, by parts (1) and (2) of the same Theorem, metrizable locally convex spaces (as in part (2) of js21's answer) and strong duals of Fréchet-Schwartz spaces also have the final topology with respect to continuous curves. This result also shows that first countability is not a necessary condition. $\endgroup$ – Pedro Lauridsen Ribeiro Jun 16 '16 at 14:43
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One can prove the following :

(1) If $X$ has enough curves, then $X$ is sequential and locally path-connected.

(2) If $X$ is first-countable and locally path-connected, then $X$ has enough curves.

Note that "first-countable" implies "sequential", but the opposite implication is not true in general.

Proof of (1) : As noted by Martin Sleziak in his first comment, if $X$ has enough curves, then it is a quotient of the disjoint sum of copies of $I$. Now, $I$ is sequential and locally path-connected, and these two properties are preserved by disjoint sums and by quotients, hence the result.

Proof of (2) : Let $f : X \rightarrow Y$ be a map sending curves to curves. We have to show that $f$ is continuous. Since $X$ is first-countable, it is sequential, hence it suffices to show that for any sequence $(x_n)_{n \geq 1}$ in $X$, converging to some $x \in X$, the sequence $(f(x_n))_{n \geq 1}$ converges to $f(x)$. Let $(U_k)_{k \geq 1}$ be a countable basis of path-connected open neighbourhoods of $x$, such that $U_k \supset U_{k+1}$ for each $k$. Wlog $x_n \in U_1$ for each $n$. For each $k \geq 1$, let $N_k$ be the smallest positive integer such that $x_n \in U_k$ for each $n \geq N_k$. In particular $N_1 = 1$. Since $U_k$ is path-connected, one can find for each $n \in [N_k, N_{k+1} [$ a continuous map $c_n : [\frac{1}{n+1}, \frac{1}{n}] \rightarrow U_k$ with $c_n(\frac{1}{n+1}) = x_{n+1}$ and $c_n(\frac{1}{n}) = x_n$. The collection $(c_n)_{n \geq 1}$ yields a continuous map $c : ]0,1] \rightarrow X$ with $c(]0,\frac{1}{N_k}]) \subseteq U_k$. In particular, setting $c(0) = x$ yields a continuous map $c : I \rightarrow X$. Since $fc$ is continuous, the sequence $f(x_n) = fc(\frac{1}{n})$ converges to $fc(0) = f(x)$.

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  • $\begingroup$ Note that "first-countable implies sequential" and (2) are each equivalent to countable choice. ​ ​ ​ (For the quote, let each $Y_n$ be the product of the first $n$ $X$s, let the [basic neighborhoods](en.wikipedia.org/wiki/Neighbourhood_system) of $\infty$ be ​ $Y_n\cup Y_{n+1}\cup Y_{n+2}...$ , ​ and let everything else in ​ $Y_0\cup Y_1\cup ...$ ​ be isolated. ​ ​ ​ For (2), disjoint union that with the natural numbers, and for each n, put in [a length-1/(2$^n$) path from n to n+1] for each element of $Y_n$.) ​ ​ ​ ​ ​ ​ ​ ​ $\endgroup$ – user5810 Jun 16 '16 at 11:04
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A space $X$ whose topology agrees with the final topology with respect to all maps $I\to X$ is often called a delta-generated ($\Delta$-generated) space. The category of $\Delta$-generated spaces is a convenient category of spaces (in the sense of Steenrod) and is used in Diffeology, directed homotopy, and generalized covering spaces theories.

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  • $\begingroup$ Let's call $\Delta^n$-generated space a space $X$ whose topology agrees with the final topology with respect to all maps $\Delta^n\to X$. To have enough curves means - if I am correct - to be $\Delta^1$-generated, so $$\Delta\mbox{-generated}\Longrightarrow \mbox{has enough curves}$$ Is it clear that the converse holds ? (It is, of course, so - by transitivity - if, for all $n$, $\Delta^n$ has enough curves.) $\endgroup$ – Duchamp Gérard H. E. Apr 9 '17 at 16:16
  • $\begingroup$ Yes, your $\Delta^n$-generated and $\Delta^1$-generated spaces are the same. This follows directly from the fact that there are space filling curves $[0,1]\to \Delta^n$ and any such map must be a topological quotient map. $\endgroup$ – Jeremy Brazas Apr 9 '17 at 16:25
  • $\begingroup$ O.K. this is what I called transitivity, but - as I am not an expert - I wanted confirmation (+1). $\endgroup$ – Duchamp Gérard H. E. Apr 9 '17 at 16:33

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