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The well-known Hahn-Mazurkiewicz theorem characterizes those nonempty Hausdorff spaces $X$ that admit a continuous surjection $\alpha: [0, 1] \to X$ from the closed unit interval: it is necessary and sufficient that $X$ be a compact, connected, locally connected metrizable space.

My Googling skills did not enable me to locate a nice way of characterizing Hausdorff spaces $X$ that admit a continuous surjection $(0, 1) \to X$ from an open interval. Is there one?

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Two thoughts: (1) every space that is the image of $[0,1]$ is the image of $(0,1)$; (2) we can replace $(0,1)$ with $\mathbb{R}$. I wonder it could be as simple as asking that $X$ be a path-connected countable union of spaces satisfying Hahn-Mazurkowicz. – Jeff Strom May 25 '14 at 1:01
One such $X$ is the Warsaw circle, which may indicate the difficulty of getting a similarly nice characterization. – Eric Wofsey May 25 '14 at 1:01
@JeffStrom I wondered the same thing. – Todd Trimble May 25 '14 at 1:01
@JeffStrom: Any path-connected countable union of Hahn-Mazurkowicz spaces can certainly be covered by a map from $(0,1)$. Simply cover each one of the Hahn-Mazurkowicz subspaces one by one, interspersed with paths between their endpoints. – Eric Wofsey May 25 '14 at 1:04
@EricWofsey: I was busy typing this up below. – Jeff Strom May 25 '14 at 1:12
up vote 8 down vote accepted

Call $A$ an HM-space if there is a continuous surjection $I\to A$ (where $I$ is the closed interval $[0,1]$). Note that if $A$ is an HM-space, then it is path-connected.

Theorem: If $X$ is path-connected, then the following are equivalent:

  1. $X = \bigcup_{n=1}^\infty A_n$ where each $A_n$ is an HM-space
  2. there is a continuous surjection $\mathbb{R} \to X$
  3. there is a continuous surjection $(0,\infty) \to X$
  4. there is a continuous surjection $[0,\infty) \to X$.

Proof: Clearly (2) and (3) are equivalent; let's prove (3) implies (1). If $f:(0,\infty) \to X$ is a continuous surjection, let $A_n = f([n-1,n])$. Then $A_n$ is evidently an HM-space and $X$ is the indicated countable union.

Now we show (1) implies (4). if $X = \bigcup_{n=1}^\infty A_n$ with each $A_n$ an HM-space, then we can choose surjective paths $\alpha_n: [2(n-1),2n-1]\to A_n$; write $x_n = \alpha_n(0)$ and $y_n= \alpha_n(1)$. Since $X$ is path-connected, we can find paths $\beta_n:[2n-1,2n]\to X$ from $y_n$ to $x_{n+1}$. Concatenating these paths gives the desired surjection $0,\infty) \to X$.

Finally, since there is a continuous surjection $\mathbb{R}\to [0,\infty )$, (4) implies (2).$\quad \square$

Just musing: there is a partial order on spaces defined by $X< Y$ if there is a continuous surjection $Y \to X$. Hahn-Mazurkowicz classifies $\{ X \mid X < I\}$. The equivalence of (2), (3) and (4) shows that half-open intervals are "equivalent" to open ones (i.e., both $(0,1)< [0,1)$ and $[0,1) < (0,1)$); and closed ones are different, being compact. Finally, note that a Peano space (i.e., HM-space) $X$ is "equivalent" in this sense to $I$ if and only if it is the domain of a nonconstant continuous function $f:X\to \mathbb{R}$.

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Well, that was easy. Thanks! That might be the best one could hope for, but I'll wait a while to see if there are other responses before accepting. – Todd Trimble May 25 '14 at 1:26
Incidentally, I think the usual term for HM-space is Peano space. – Todd Trimble May 25 '14 at 1:29
Just to clarify, $I$ here denotes the closed unit interval? – Nate Eldredge May 25 '14 at 1:39
@NateEldredge That's right. – Todd Trimble May 25 '14 at 1:45
This characterization appears in (see Proposition 3.2). The author calls these spaces $\sigma$-Peano spaces. – Ramiro de la Vega May 28 '14 at 14:45

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