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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

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    $\begingroup$ 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 $*$. $\endgroup$ – Mark Kamsma May 21 at 14:35
  • $\begingroup$ map any discrete space onto a singleton $\endgroup$ – Robert Thingum May 21 at 15:34
  • $\begingroup$ 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. $\endgroup$ – PrimeRibeyeDeal May 21 at 16:24
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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.

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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.

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