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Whenever $X$ is a topological space, let us say that continuity on $X$ is detected by $\mathbb{R}$ iff for all functions $f : X \rightarrow Y$ where $Y$ is another topological space, we have that if every continuous function $c : \mathbb{R} \rightarrow X$ satisfies "$f \circ c$ is continuous", then $f$ is itself continuous.

  • Clearly, continuity on $\mathbb{R}$ is detected by $\mathbb{R}$
  • It is shown here that continuity on $\mathbb{R}^2$ is detected by $\mathbb{R}$.
  • Clearly, it is not the case that continuity on $\mathbb{Q}$ is detected by $\mathbb{R}$, since every continuous mapping $\mathbb{R} \rightarrow \mathbb{Q}$ is constant.

Question. Do there exist general conditions underwhich we can conclude that continuity on a topological space is detected by $\mathbb{R}$?

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  • $\begingroup$ maybe, locally path-connected? $\endgroup$ – Andrey Ryabichev Aug 22 '15 at 18:31
  • $\begingroup$ I think locally path connected $\endgroup$ – Anubhav Mukherjee Aug 22 '15 at 18:33
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    $\begingroup$ A locally path-connected first countable space has that property $\endgroup$ – Daniel Fischer Aug 22 '15 at 18:38
  • $\begingroup$ This comment indicates that $X$ has the final topology with respect to all maps from open subsets of some Euclidean space (which is called there numerically generated) iff it has the final topology with respect to all map $I\to X$, which is think is the final topology of all maps $\Bbb R\to X$. $\endgroup$ – Stefan Hamcke Aug 22 '15 at 18:53
  • $\begingroup$ @DanielFischer, neat. So every first-countable locally-euclidean space has this property; thus, in particular, every topological manifold has it. $\endgroup$ – goblin GONE Aug 22 '15 at 18:56
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Let us say a space $X$ is $\mathbb{R}$-generated if continuity of maps out of $X$ is detected by composition with maps $\mathbb{R}\to X$. By general nonsense, this is the same as saying a subset of $X$ is closed (or open) iff its preimage under every continuous map $\mathbb{R}\to X$ is, or that $X$ is the colimit of a diagram in which every object is a copy of $\mathbb{R}$.

Consider the following statements:

  1. $X$ is locally path-connected and first-countable.
  2. $X$ is $\mathbb{R}$-generated.
  3. $X$ is locally path-connected and sequential.

I claim that $(1)\Rightarrow (2)\Rightarrow (3)$, and neither implication is reversible.

To show $(1)\Rightarrow (2)$, suppose $X$ is locally path-connected and first-countable. It suffices to show that if a subset $A\subseteq X$ is not closed, then there is some continuous $f:\mathbb{R}\to X$ such that $f^{-1}(A)$ is not closed. So suppose $A\subseteq X$ is not closed; let $x\in\bar{A}\setminus A$. Let $(U_n)$ be a local neighborhood base at $x$; let $x_n\in U_n\cap A$ for each $n$. Choose a path from $x_n$ to $x_{n+1}$ that lies entirely inside $U_n$ for each $n$. Concatenating these paths together, you get a map $f_0:[0,1)\to X$ which extends continuously to $[0,1]$ by sending $1$ to $x$. Extending this to a map $f:\mathbb{R}\to X$, we then have that $f^{-1}(A)$ contains a sequence that converges to $1$ but does not contain $1$, and so $f^{-1}(A)$ is not closed.

To show $(2)\Rightarrow (3)$, suppose $X$ is $\mathbb{R}$-generated. Since $\mathbb{R}$ is sequential, $X$ must be sequential. Now enlarge the topology on $X$ by saying that if $U$ is an open neighborhood of a point $x\in X$, so is the path-component of $x$ inside $U$. Let $Y$ be $X$ with this enlarged topology; it is easy to see that any continuous map $\mathbb{R}\to X$ is still continuous as a map $\mathbb{R}\to Y$. Since $X$ is $\mathbb{R}$-generated, this means the identity map $X\to Y$ is continuous. That is, our enlarged topology is the same as the original topology; it follows that $X$ is locally path-connected.

Finally, let us give some counterexamples to the reverse implications. For $(2)\not\Rightarrow(1)$, note that a colimit of $\mathbb{R}$-generated spaces is $\mathbb{R}$-generated, so any CW-complex is $\mathbb{R}$-generated. But a CW-complex is first-countable iff it is locally finite.

A counterexample to $(3)\Rightarrow(2)$ can be obtained as follows. Let $W=\mathbb{N}\times(0,1]\cup\{\infty\}$ be the 1-point compactification of $\mathbb{N}\times(0,1]$, and let $Z$ be the subspace $\mathbb{N}\times(0,1)\cup\{\infty\}\subset W$. It is easy to see that $Z$ is locally path-connected and sequential. Let $X=\mathbb{N}\times Z\cup\{x\}$, topologized as follows. Any open subset of $\mathbb{N}\times Z$ (with the product topology) is open in $X$, and a set $U$ containing $x$ is open iff the following conditions hold:

  • $U\cap\mathbb{N}\times Z$ is open in $\mathbb{N}\times Z$.
  • For every $m,n\in\mathbb{N}$, there is a $t_{m,n}<1$ such that $\{m\}\times\{n\}\times(t_{m,n},1)\subset U$.
  • For all but finitely many $m\in\mathbb{N}$, $(m,\infty)\in U$.

For any $m,n\in\mathbb{N}$, the map $f:[0,1]\to X$ such that $f(0)=(m,\infty)$, $f(t)=(m,n,t)$ for $0<t<1$, and $f(1)=x$ is continuous. It follows that $X$ is locally path-connected. To show that $X$ is sequential, it suffices to show that if $A\subseteq\mathbb{N}\times Z$ is closed in $\mathbb{N}\times Z$ and $x\in\bar{A}$ in $X$, then some sequence in $A$ converges to $x$. If there exist $m,n\in \mathbb{N}$ such that $(m,n,t)\in A$ for $t$ arbitrarily close to $1$, a sequence of such points with $t\to 1$ will converge to $x$. If no such $m$ and $n$ exist, then there must either be infinitely many $m$ such that $(m,\infty)\in A$ or infinitely many $m$ such that for infinitely many $n$, $(m,n,t)\in A$ for some $t\in(0,1)$. Since $A$ is closed in $\mathbb{N}\times Z$, the second case actually implies the first case (since in $Z$, any sequence of points $(n,t)$ with $n\to\infty$ converges to $\infty$). Thus we can find a sequence of points $(m,\infty)\in A$ with $m\to \infty$, and such a sequence converges to $x$.

Finally, I claim $X$ is not $\mathbb{R}$-generated. Indeed, consider the set $A=\mathbb{N}\times\{\infty\}\subset X$. It is easy to see that $\bar{A}=A\cup\{x\}$, but I claim that $f^{-1}(A)$ is closed for every continuous $f:\mathbb{R}\to X$. To show this, let $f:\mathbb{R}\to X$ be continuous; then the only way $f^{-1}(A)$ can fail to be closed is if some sequence in $f^{-1}(A)$ converges to a preimage of $x$. Thus suppose (WLOG) that $f(0)=x$ and $f(1/k)=(m_k,\infty)\in A$ for each positive integer $k$. If there is some $m$ such that $m_k=m$ for infinitely many $k$, then clearly $f$ will fail to be continuous at $0$, so we may assume the $m_k$ are all distinct. Note that any path in $X$ from $(m,\infty)$ to $(m',\infty)$ must pass through $x$ if $m\neq m'$, and that any path from $(m,\infty)$ to $x$ must pass through $\{m\}\times\mathbb{N}\times(0,1)$. It follows that for each $k$, there is some $s_k\in(1/k,1/(k+1))$ such that $f(s_k)\in \{m_k\}\times\mathbb{N}\times(0,1)$. Since $f$ is continuous, the sequence $f(s_k)$ must converge to $f(0)=x$. But since the $m_k$ are all distinct, it is easy to find a neighborhood of $x$ that does not contains any $f(s_k)$. This contradiction shows that $f^{-1}(A)$ is closed. Since $f$ was arbitrary, this shows $X$ is not $\mathbb{R}$-generated.

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  • $\begingroup$ Thanks. I'll take a better look at your answer when I have more time to spare. But, why the terminology "$W$-generated"? In category theory, I think this phrase tends to mean something else, namely: $X$ is $W$-generated iff for any two parallel outbound morphisms $f,g : X \rightarrow Y$, we have that if $f \circ x = g \circ x$ for all morphisms $x : W \rightarrow X$, then $f=g$. $\endgroup$ – goblin GONE Aug 23 '15 at 3:20
  • $\begingroup$ This use of "generated" is consistent with other uses in pointset topology, such as "compactly generated" or "countably generated". It is also similar to a usage sometimes found in category theory (particularly in the context of things like triangulated categories), where "generated" means something like "generates using colimits" (or, as is equivalent under appropriate hypotheses, "detects isomorphisms"). $\endgroup$ – Eric Wofsey Aug 23 '15 at 11:39
  • $\begingroup$ Honestly, I think its best to use slightly longer phrases for these concepts. Like "Suppose continuity on $X$ is $\mathbb{R}$-detected..." or "Suppose closedness in $X$ is compactly detected..." or "Suppose closedness in $X$ is countably detected." etc. Sure, its a bit long-winded, but I really do think this way of talking is ultimately clearer and more conducive to being understood by a wider audience. $\endgroup$ – goblin GONE Aug 23 '15 at 13:41
  • $\begingroup$ I mean, I wouldn't just start using a term like "$\mathbb{R}$-generated" without defining it (unless it was obvious in context). I'm not convinced "continuity on $X$ is $\mathbb{R}$-detected" is actually significantly clearer to someone who doesn't know the context of what you're talking about anyways (I would not consider my opening sentence in which I define "$\mathbb{R}$-generated" an adequate definition without the context provided by the question). $\endgroup$ – Eric Wofsey Aug 23 '15 at 13:57
  • $\begingroup$ But I also think "compactly generated" is confusing terminology. This could easily be interpreted to mean: "has a dense subset that is compact." Same goes for "countably generated." Just my 2 cents. $\endgroup$ – goblin GONE Aug 23 '15 at 14:26

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