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Questions. Let 'arc in a topological space'$:=$'map $[0,1]\rightarrow X$ which is a homeomorphism onto its image'.

  1. What do you consider usual citable references for the following statement?

Let $X$ be a Hausdorff space. Let $x_0,x_1,x_2\in X$ be mutually distinct. For both $i\in 2$ let $[0,1]\xrightarrow[]{A_i}X$ be an arc with $A_i(0)=x_i$ and $A_i(1)=x_{i+1}$. Then there exists an arc $[0,1]\xrightarrow[]{A_2}X$ with $A_2(0)=x_0$ and $A_2(1)=x_2$.

  1. What is the weakest separation axiom (known to you) on $X$ other the above-mentioned $\textsf{T}_2$ that makes the above conclusion true, and what are references for/discussions of the significance of, such more general 'arc-composition-theorem'?

Remarks.

  • Without mutual distinctness of the points, $x_0=x_2=0\in\mathbb{R}$, $x_1=1$, $X=\mathbb{R}$ and $A_1$ and $A_2$ e.g. the linearly-parametrized arcs, is a counterexample.
  • Motivation for this question is working on problems about infinite planar (di)graphs.
  • related
  • related
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  • $\begingroup$ That statement is not always true. For a counterexample, take $X=\mathbb{R}$, $x_0=x_2=0$, $x_1=1$ (wasn't sure what $i \in 2$ means, so I replaced it in my mind with $i \in \{0,1\}$). $\endgroup$ – Lee Mosher Aug 22 '17 at 15:01
  • $\begingroup$ @LeeMosher: you are right. Thanks for that. Hopefully you agree that the statement is 'morally right'. The problem that you kindly point out is that with my careless definition there is no 'identity arc'. For this strong reason alone, 'arcs' in the usual sense do not form a category. One might speculate that this is why while there is a very standard tool called 'path category', there is no noticeable notion of 'arc category'. I will simply patch the question by requiring the points to be distinct. $\endgroup$ – Peter Heinig Aug 22 '17 at 15:20
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I'm afraid I don't know of any references, but I hope the following discussion may nevertheless be helpful to you.

First, there is a very simple proof of this fact, so simple that I wouldn't be surprised if there is no standard reference. You just truncate $A_0$ at the first point where it crosses $A_1$, and then concatenate the path $A_0$ up to that point with the remainder of $A_1$. In detail: consider the set $$C=\{(s,t)\in [0,1]\times[0,1]:A_0(s)=A_1(t)\}.$$ Since $X$ is Hausdorff, $C$ is closed (it is the inverse image of the diagonal in $X\times X$ under the map $(A_0,A_1)$). Thus $C$ is compact, and so its projection $$D=\{s\in[0,1]:A_0(s)=A_1(t)\text{ for some }t\in [0,1]\}$$ is also compact (and nonempty since $A_0(1)=A_1(0)$). Let $s$ be the least element of $D$, and let $t\in [0,1]$ be such that $A_0(s)=A_1(t)$. Now simply define $A_2$ to be the concatenation of the paths $A_0|_{[0,s]}$ and $A_1|_{[t,1]}$. This path $A_2$ will be injective by minimality of $s$, and is thus an arc since any continuous injection from a compact space to a Hausdorff space is an embedding. Since $x_0\neq x_2$, we have either $s>0$ or $t<1$, so the domain of the concatenation really is a nondegenerate interval and not just a single point.

Second, I think Hausdorffness is pretty much the only natural hypothesis to require here, and would be surprised if there is any interestingly weaker separation axiom that suffices. I think the following example of a non-Hausdorff space where it fails is illustrative. Let $X$ be $[0,1]$ with the endpoint $0$ "doubled": that is, $X=[0,1]\cup\{0'\}$ where neighborhoods of $0'$ are sets of the form $\{0'\}\cup(0,\epsilon)$. There is an arc $A_0$ from $0$ to $1$ and an arc $A_1$ from $1$ to $0'$, namely $A_0(t)=t$ and $A_1(t)=1-t$ for $t<1$ and $A_1(1)=0'$. But there is no arc from $0$ to $0'$, since any nonconstant path starting at $0$ must traverse all of $(0,\epsilon)$ for some $\epsilon>0$ and then in order to reach $0'$ it must pass back through the same interval.

The only separation axiom I can think of which is weaker than Hausdorff and suffices is that $X$ be weakly Hausdorff, meaning that the image of any continuous map from a compact Hausdorff space to $X$ is closed. This suffices, since in that case the image of the concatenation of $A_0$ and $A_1$ is Hausdorff (the image of any map from a compact Hausdorff space to $X$ is Hausdorff; see Lemma 1.4(b) of http://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf), and so you can just restrict your attention to that image.

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