# Continuous if and only if graph of every path connected subspace is path connected?

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.

• 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. – bof Dec 19 '20 at 5:53
• You’re going to want to make some assumptions about $X$ and $Y$; at the very least locally path-connected, maybe even manifolds. – Qiaochu Yuan Dec 19 '20 at 6:40
• @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 $X = Y \mathbb{R}$. – Charles Hudgins Dec 19 '20 at 14:40

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.

• 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 $\mathbb{R}$. – Charles Hudgins Dec 19 '20 at 14:45