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It is clear that if $f : X \to Y$ is continuous then the graph of $f$ over every path connected subspace is path connected.

Is the converse true? That is, if the graph of $f : X \to Y$ over every path connected subspace is path connected, does it follow that $f$ is continuous?

The topologist's sine curve is a well known counterexample if we say "connected" instead of "path connected." Is there a standard counterexample in the path connected case? Is there a counterexample that works when $X = Y = \mathbb{R}$?

Part of why I ask is that continuity is often informally defined in introductory calculus classes as a function whose graph you could "walk along" as though it were a path. This, however, is quite different from the formal definition and, I suspect (but have not been able to prove), is not even true.

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  • $\begingroup$ What are you assuming about $X$? I guess there will be some easy counterexamples where $X$ is a non-discrete space with no nontrivial path-connected subspaces, e.g., the space of irrational numbers. $\endgroup$ – bof Dec 19 '20 at 5:53
  • $\begingroup$ You’re going to want to make some assumptions about $X$ and $Y$; at the very least locally path-connected, maybe even manifolds. $\endgroup$ – Qiaochu Yuan Dec 19 '20 at 6:40
  • $\begingroup$ @bof It should have been obvious to me that we needed more assumptions on $X$ and $Y$. I'm really most interested in the case where $X = Y \mathbb{R}$. $\endgroup$ – Charles Hudgins Dec 19 '20 at 14:40
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For arbitrary spaces it certainly isn't true. For instance, if $X$ is totally path-disconnected (e.g., $X=\mathbb{Q}$), then your condition is trivial, but not every map out of $X$ is continuous unless $X$ is discrete.

To characterize the spaces for which it is true, consider the following properties of a topological space $X$:

  • $X$ is well-graphed if it has your property for arbitrary $Y$: that is, a map $X\to Y$ is continuous iff its graph over every path-connected subspace of $X$ is path-connected.
  • $X$ is path-generated if for any $Y$, a map $X\to Y$ is continuous iff its composition with every continous path $[0,1]\to X$ is continuous. (For more about this property see this answer of mine; in particular, a path-generated space must be locally path-connected and sequential.)
  • $X$ is arc-generated if for any $Y$, a map $X\to Y$ is continuous iff its composition with every embedding $[0,1]\to X$ is continuous.

We then have the following theorem:

Theorem: Every arc-generated space is well-graphed, and the converse holds for Hausdorff spaces. Every well-graphed space is path-generated.

Proof: Suppose $X$ is well-graphed and $f:X\to Y$ has the property that the composition of $f$ with every path in $X$ is continuous. Then the graph of $f$ over every path-connected subspace is path-connected, since every path in $X$ lifts to a path in the graph. Thus $f$ is continuous, so $X$ is path-generated. Moreover, if $X$ is Hausdorff, then it would suffice to assume the composition of $f$ with every embedding $[0,1]\to X$ is continuous, since in a path-connected Hausdorff space any two points are connected by a path which is an embedding, so $X$ would be arc-generated.

Conversely, suppose $X$ is arc-generated. Let $f:X\to Y$ be a map whose graph over every path-connected subset is path-connected. To show $f$ is continuous, it suffices to show $fg$ is continuous for every embedding $g:[0,1]\to X$. Since $g$ is an embedding, the graph of $fg:[0,1]\to Y$ is homeomorphic to the graph of $f$ restricted to the image of $g$. By hypothesis, the latter graph is path-connected, so the graph of $fg$ is path-connected. Now take a path $h$ from $(0,f(g(0)))$ to $(1,f(g(1)))$ in this graph. The first coordinate of $h$ is a continuous surjection $p:[0,1]\to[0,1]$, which is automatically a quotient map since the domain is compact and the codomain is Hausdorff. The second coordinate of $h$ is then the map $fgp:[0,1]\to Y$ and is continuous since $h$ is. But since $p$ is a quotient map, this implies $fg$ is continuous, as desired. $\blacksquare$

In particular, $\mathbb{R}$ is arc-generated and thus well-graphed, so your statement is true for the motivating example of maps $\mathbb{R}\to\mathbb{R}$.

As a counterexample to show that a Hausdorff path-generated space need not be well-graphed (or equivalently, arc-generated), let $X$ be the 1-point compactification of the set $$A=[0,\infty)\times\{0\}\cup\mathbb{N}\times[0,1]\subset\mathbb{R}^2.$$ Then $X$ is path-generated (indeed, it is a quotient of $[0,1]$, by a path that goes from $(0,0)$ off to $\infty$ taking a detour on each $\{n\}\to[0,1]$ along the way). However, I claim $X$ is not well-graphed. Consider the map $f:X\to[0,1]$ that is the first projection on $A$ and maps $\infty$ to $0$. Then $f$ is not continuous at $\infty$, since the points $(n,1)\in A$ approach $\infty$ as $n$ gets large but $f(n,1)=1$ and $f(\infty)=0$. However, the graph of $f$ over any path-connected set is path-connected, since any path-connected subset that contains both $\infty$ and a point $(n,t)$ for $t>0$ must also contain the path from $(n,t)$ down to $(n,0)$ and then the horizontal path from $(n,0)$ out to $\infty$, and $f$ is continuous on both of these paths.

I don't know whether there is a (necessarily non-Hausdorff) well-graphed space that is not arc-generated.

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  • $\begingroup$ My question could have been phrased better, but this answers everything. Thank you. I'm a bit surprised that turning "connected" into "path-connected" makes the statement true for $\mathbb{R}$. $\endgroup$ – Charles Hudgins Dec 19 '20 at 14:45

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