TITLE: Which spaces have enough curves
QUESTION [14 upvotes]: 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?

REPLY [8 votes]: 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 $fc(0) = f(x)$.

REPLY [7 votes]: 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.