# Topological space which is not path connected but has a continuous surjection on a Space which is path connected

I am trying to learn a bit about topology and I've found a problem, where I have to construct a topological space which is not path connected but has a continuous surjection on a space which is path connected.

My idea was that I could take connected space which is not path connected and map every open set on a single point. This single point should be path connected and the mapping fulfills the requirement.

However, I am not sure if this is right and I need a few hints

• Your solution is pretty much correct. Only you should not define your map on open sets, but just on elements of your space. So take your favourite disconnected space $X$ (or connected, but not path-connected, if you wish) and consider $Y = \{ * \}$. Then define $f: X \to Y$ by sending every $x \in X$ to $*$. – Mark Kamsma May 21 '19 at 14:35
• map any discrete space onto a singleton – Robert Thingum May 21 '19 at 15:34
• An interesting such example of a connected space is the natural surjection of the solenoid onto the circle $S^1$. The solenoid is an inverse limit of circles, but is itself only connected and not path connected. It's even a topological group. – PrimeRibeyeDeal May 21 '19 at 16:24

Just take $$X=[0,1]\times \{ 1,2\}$$ and send it to $$[0,1]$$ by $$\pi:X\to [0,1]$$ given by $$\pi(x,n)= x$$. This is a continuous surjection, but the domain is disconnected while the image is path connected.
Let $$X$$ be your favourite path-connected space and let $$Y$$ be the same set in the discrete topology (which is totally disconnected if $$|X|=|Y| \ge 2$$). Then $$f(x)=x$$ is continuous from $$Y$$ onto $$X$$. Any function on a discrete space is continuous.