2

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 $card(X)\geq card(Y)$.

Eric Wofsey
  • 342,377
Emo
  • 3,719

2 Answers2

7

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 $Y\to X$).

Alessandro Codenotti
  • 12,895
  • 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 Aug 21 '23 at 12:31
  • 1
    @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. – Alessandro Codenotti Aug 21 '23 at 17:23
  • 1
    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 – Alessandro Codenotti Aug 21 '23 at 17:24
6

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.

Eric Wofsey
  • 342,377
  • Elegant example! May I know the reference or is it originaly yours? – Emo Aug 22 '23 at 08:19
  • 1
    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). – Eric Wofsey Aug 22 '23 at 13:16