# Do any two connected spaces have a continuous surjection between them?

It is known that if $$X$$ is a connected topological space and there exists a continuous surjection $$f:X\to Y$$, then so is $$Y$$.

I wonder if there exist connected topological spaces $$X$$ and $$Y$$ such that there is no continuous surjection between them? I first thought of $$S^1$$ and Warsaw circle, but it seems that it is not hard to construct a contiuous surjection from Warsaw circle to $$S^1$$ by cutting the Warsaw circle into infinitely many intervals and mapping each of them to $$S^1$$.

Remark: WLOG we take $$X$$ and $$Y$$ such that $$card(X)\geq card(Y)$$.

• Hint: if $X$ is compact and $Y$ is Hausdorff, then $Y$ has to be compact for $f$ to exist. Commented Aug 21, 2023 at 10:55
• What if $X$ is a one point space, and $Y$ is a larger connected set?
– Mark
Commented Aug 21, 2023 at 10:57
• @Mark I can construct a continuous surjection from $Y$ to $X$.
– Emo
Commented Aug 21, 2023 at 11:01

Let $$X$$ be a path connected space and $$Y$$ be a connected space which is not path connected, with $$|X|>|Y|$$. (for example you can take $$Y$$ to be the topologists's sine curve and $$X=\Bbb R^\kappa$$ for big enough $$\kappa$$) Then there is no continuous surjection $$Y\to X$$, because there are no surjections $$Y\to X$$ at all, while there is no continuous surjection $$X\to Y$$ because the continuous image of a path connected space is path connected.

There is a positive result along the lines of your question if you work with nice spaces: suppose $$X$$ and $$Y$$ are compact connected metrizable locally connected spaces (also known as Peano continua) which are not singletons. Then there is a continuous surjection $$X\to Y$$ (and a continuous surjection $$Y\to X$$).

• What's the nature of $\kappa$ on this case? I suppose $\kappa$ must not be a positive integer. I am not familiar with these spaces, could you provide us with some references?
– Emo
Commented Aug 21, 2023 at 12:31
• @Emo $\kappa$ is some arbitrary infinite cardinal which is big enough so that $|\Bbb R^\kappa|>|\Bbb R|$ (since the latter is the cardinality of the topologist's sine curve ). $\Bbb R^\kappa$ is path connected (in the product topology) since it is a product of path connected spaces. Commented Aug 21, 2023 at 17:23
• But note that the specific spaces don't matter, all you need for the argument to work is $Y$ which is connected but not path connected and $X$ which is path connected with $|X|>|Y|$, you can use your favourite spaces that have those properties instead of those from my answer Commented Aug 21, 2023 at 17:24

Here is perhaps the simplest example (or at least, the smallest possible example!). Let $$X=Y=\{a,b,c\}$$ with the following topologies. The open sets in $$X$$ are $$\emptyset,\{a\},\{b\},\{a,b\},\{a,b,c\}$$. The open sets in $$Y$$ are the closed sets of $$X$$: $$\{a,b,c\},\{b,c\},\{a,c\},\{c\},\emptyset$$. Since $$X$$ and $$Y$$ have the same finite cardinality and the same number of open sets, any continuous surjection between them must be a homeomorphism. However, they are not homeomorphic since $$X$$ has two open singletons and $$Y$$ has only one.

• Elegant example! May I know the reference or is it originaly yours?
– Emo
Commented Aug 22, 2023 at 8:19
• I came up with it. It's easy to come up with examples like this if you know some general theory of finite topological spaces (see here for some relevant ideas). Commented Aug 22, 2023 at 13:16