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Under what circumstances is there at least one non-constant continuous function from a topological space $X$ to a topological space $Y$? Assume that $X$ and $Y$ each have at least two points. If $X$ is disconnected, separated by $A$ and $B$, then any function with one value on $A$ and another on $B$ is continuous. If $X$ is connected, then the image of $X$ under a continuous function must lie within a connected component of $Y$. Therefore, to avoid triviality, assume that $X$ and $Y$ are both connected.

The only theorem I've encountered of this nature is Urysohn's lemma, which proves such a function exists if $X$ is a $T_4$ space and $Y$ has a path-connected component with more than one point. This is of course a rather strong condition.

It's obvious that if $X$ is convex in $\mathbb{R}$ and $Y$ is totally path disconnected, then there is no such function.

Otherwise, I haven't a clue. I'm particularly curious about what happens if $X$ and/or $Y$ is required to be homogeneous or bihomogeneous, and/or if $Y$ is required to be uniform.

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@t.b. That is indeed what I meant. –  dfeuer Oct 13 '11 at 23:20
    
This paper might be of interest, although it does not answer your question. Horst Herrlich: Wann sind alle stetigen Abbildungen in Y konstant? Mathematische Zeitschrift, Volume 90, Number 2, 152-154 (Probably can be found also at GDZ - gdz.sub.uni-goettingen.de/dms/load/toc/?IDDOC=8487. But the GDZ link is not working for me at the moment.) –  Martin Sleziak Nov 1 '11 at 12:26
    
@Martin Sleziak Unfortunately, I know absolutely no German, so I can't read that. Do you know of an English translation? –  dfeuer Nov 17 '11 at 20:59
    
I've put a translation of the main result of that paper into an answer bellow. –  Martin Sleziak Nov 17 '11 at 21:13

2 Answers 2

up vote 10 down vote accepted

This paper might be of interest:

The content of the paper:

Urysohn [5] asked whether for every regular space $X$ (having at least two points) there is a non-constant continuous map from $X$ do the space $Y$ of real numbers. This question was negatively answered by Hewitt [2], Novak [2] and Van Est-Freudenthal [1]. The methods used by these authors (which go back to Tychonoff [4]) let us show relatively easy the following result:

Theorem Let $Y$ be a topological space. The following conditions are equivalent:
(a) $Y$ is a $T_1$-space,
(b) there exists a regular space $X$ (having at least two points), such that every continuous map from $X$ to $Y$ is constant.

Sketch of the construction in the proof of this theorem: (I have omitted many details and also the proofs that these space do have the required properties.)

Definition of a space $Q$. First we start with some given space $Y$.

  • The spaces $R_i$ for $i=1,2$ and points $r_i\in R_i$ are constructed in such way that every continuous map from $R_i$ to $Y$ is constant on some neighborhood of $R_i$.
  • A space $T=R_1\times R_2\setminus \{(r_1,r_2)\}$. This space has the property, that for every continuous map $f$ from $T$ to $Y$ there exist neighborhoods $U_i$ of $r_i$ such that $f$ is constant on $U_1\times U_2 - \{(r_1; r_2)\}$.
  • We take countably many homeomorphic copies $T\times\{n\}$ of the space $T$. We add two new points $a$, $b$ with local neighborhood bases $\{\bigcup T_m; m\ge n\}\cup \{a\}\subseteq B$ and $\{\bigcup T_m; m\ge n\}\cup \{b\}\subseteq B$.
  • In this space we identify $(x,r_2,n)$ and $(x,r_2,n+1)$ for any $x\in R_1\setminus\{r_1\}$ and any even $n$. We also identify $(r_1,x,n)$ and $(r_1,x,n+1)$ for every odd $n$ and every $x\in R_2\setminus\{r_2\}$.

Let us call the resulting space $Q$.

Now for any space $Z$ we define a space $Q(Z)$ on the set $Z\times Q$ where a subset $B$ of $Z\times Q$ is open in $Z\times Q$ if and only if the following holds:

  • If $(z;x)$ is an element of $B$, then there is a neighborhood $U$ of $x$ in $Q$ with $\{z\}\times U\subset B$.
  • If $(z;a)$ is an element of $B$, then there is a neighborhood $U$ of $z$ in $Z$ with $U\times\{a\}\subset B$.

If we identify in the above space $Z\times Q$ all points of the set $Z\times\{b\}$, then we obtain a space $Q(Z)$.

Definition of $Q(Z)$. The space $Q(Z)$ contains a homeomorphic copy of $Z$. If $f$ is a continuous map from $Q(Z)$ to $Y$, then $f$ is constant on $Z$.

Definition of $X$. Let $X_0$ be a singleton. By induction we define $X_{n+1}=Q(X_n)$. Then $X_0\subset X_1\subset X_2\subset \dots $ are regular spaces. Let a subset of $X=\bigcup\{X_n; n=0,1,\dots\}$ be open if and only if $B\cap X_n$ is open for every $n$. The space $X$ is a regular space. Every continuous map from $X$ to $Y$ is constant.

Remark. The above results has a trivial analogue:
Let $X$ be a topological space. The following conditions are equivalent:
(a) $X$ is connected,
(b) there is a regular space $Y$ (having at least 2 points), such that every continuous map from $X$ to $Y$ is constant.

EDIT: I have put my attempt to translate the article here (let me know if you find any typos or mistranslations).

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"every constant map from X to Y is constant" –  Damian Sobota Nov 18 '11 at 1:03
    
@Damian. I've corrected the typo. thanks for noticing. –  Martin Sleziak Nov 18 '11 at 8:16
    
@Martin. You corrected on of the two occurrences of this typo. –  Johan Nov 18 '11 at 11:30
    
@Johan Thanks. I really should be more careful when typing. –  Martin Sleziak Nov 18 '11 at 11:34
    
@Martin. Good one. Thanks for the partial translation. –  dfeuer Nov 19 '11 at 16:46

I doubt that a complete and simple characterization exists. One common obstruction though (generalizing a bit what you said) is if $X$ is connected and $Y$ is completely disconnected.

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