Let $X$ be a topological space, let $U \subset X$, and suppose that for every path $\gamma\colon [0,1] \to X$ the preimage $\gamma^{-1}(U)$ is open. Is it true that $U$ is open? Presumably not in general, but are there reasonable requirements we can put on $X$ to make it true?

To put it another way, the subsets $U$ as above give a topology on $X$, which is the same as or finer than the original topology. When is it the same?

To put it a third way, consider the evaluation map $I \times C(I,X) \to X$, where $I = [0,1]$ and $C(I,X)$ is the set of continuous maps, with (unusually) the discrete topology. When is this is a quotient map?

  • 4
    $\begingroup$ The $p$-adics (or any non-discrete, totally disconnected space) admit no non-constant paths. $\endgroup$
    – LSpice
    Feb 4, 2020 at 2:48
  • $\begingroup$ Nice. And the topologist's sine curve will become disconnected in this topology, so maybe we want to say locally path-connected at least. $\endgroup$ Feb 4, 2020 at 2:51
  • 4
    $\begingroup$ Any topology on $[0,1]$ strictly coarser than the usual one gives a counterexample - and e.g. the cofinite topology on $[0,1]$ is path connected. I think the right context for this question, at least at first, is (locally) path-connected Hausdorff spaces. $\endgroup$ Feb 4, 2020 at 3:10
  • 3
    $\begingroup$ The spaces satisfying this condition are often called “delta generated” spaces. $\endgroup$ Feb 4, 2020 at 3:42
  • $\begingroup$ What about locally path connected spaces? $\endgroup$ Feb 4, 2020 at 5:06

1 Answer 1


A space $X$ is called $\Delta$-generated if $U$ is open in $X$ if and only if $\alpha^{-1}(U)$ is open in $[0,1]$ for every path $\alpha:[0,1]\to X$.

It's easy to see that a space $X$ is $\Delta$-generated if and only if $X$ is a quotient space of a disjoint union of copies of $[0,1]$. It follows that every $\Delta$-generated space is a locally path-connected $k$-space. It's also not too hard to show that every first countable locally path-connected space is $\Delta$-generated. Hence,

$$\text{locally path connected & first countable}\Rightarrow \Delta\text{-generated}\Rightarrow\text{locally path connected & }k\text{-space}$$

The category of $\Delta$-generated spaces is very well-behaved, namely, it is a Cartesian closed coreflective subcategory of $\mathbf{Top}$.

The following ``directed hedgehog space" is a locally path-connected space that is not $\Delta$-generated: Let $(A_{\lambda},a_{\lambda})$ be a copy of the based unit interval $([0,1],0)$ for all ordinals $\lambda<\omega_1$. Consider the one-point union $X=\bigvee_{\lambda<\omega_1}A_{\lambda}$ with based point $x_0$ and with the directed wedge topology. This means that $U$ is open in $X$ if $U\cap A_{\lambda}$ is open in $A_{\lambda}$ for all $\lambda$ and if when $x_0\in U$, there exists $\kappa<\omega_1$ such that $A_{\lambda}\subseteq U$ for all $\lambda\geq \kappa$. Clearly, $X$ is locally path-connected and as the quotient of a compact Hausdorff space, $X$ is a $k$-space. Let $b_{\lambda}$ be the image of $1$ in $A_{\lambda}$, i.e. the end of the $\lambda$-th arc, and set $C=\{b_{\lambda}\mid \lambda<\omega_1\}$. Since $[0,1]$ is sequentially compact and no sequence of countable ordinals converges to $\omega_1$, one can show that every path in $X$ intersects $C$ in at most finitely many points. Even though $C$ is not closed in $X$, $\alpha^{-1}(C)$ is closed in $X$ for every path $\alpha:[0,1]\to X$.

By similar reasoning, the quotient space $\displaystyle X'\cong \frac{(\omega_1+1)\times [0,1]}{\{\omega_1\}\times [0,1]\cup (\omega_1+1)\times\{0\}}$ will be a non-$\Delta$-generated $k$-space but it is not locally path-connected. I imagine that you can build a non-$\Delta$-generated space that is both locally path-connected and a $k$-space, but I don't know one off the top of my head. The $k$-space coreflection of $X$ and locally-path connected coreflection of $X'$ don't seem to work.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.