# Is there a continuous function from X onto Y?

(1) $X= (0,1]$ and $Y = [0,1]$;

(2) $X = [0,1]$ and $Y$ is the topologist's sine curve

(3) $X = [0,1] \cup [2,3]$ and $Y = X \times X$

I believe there is a continuous function for the first one.

I know there doesn't exist a continuous function for the second one, because $X$ is path connected and the topologist's sine curve is not. As well as, $f(0)$ doesn't exist in $Y$.

I feel like there is no continuous function for (3) because of the disconnection; however, I could be wrong.

• Count connected components. – Daniel Fischer Apr 25 '14 at 18:44
• What topology do the sets have? A function $\mathrm{f} : X \to Y$ is called continuous if the pre-image of each open set in $Y$ is an open set in $X$. To know if this is true, or not, you need to know which sets are open. You need to state the topologies that you are using on each choice of $X$ and $Y$. For example, if $X$ is given the discreet topology then every function is continuous. – Fly by Night Apr 25 '14 at 19:00
• Well, you can take $f(x)=|2x-1|$ (or some similar piecewise linear function) to get a surjective continuous map from $(0,1]$ onto $[0,1]$. – Martin Sleziak Apr 25 '14 at 20:38

1. $Y$ has the property that if you remove two certain points (namely $0$ and $1$), it is still connected. Does $X$ has this property?
2. Are $X$ and $Y$ path-connected?
3. How many connected components does $X$ have? What about $Y$?
• That's right. So if there were a surjective continuous function $X\to Y$ we could take any path in $X$ and map it over to get a path in $Y$. Do you see a contradiction? – user134824 Apr 26 '14 at 18:36