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I have drawn the two different functions and I can fully see that $|\pi_0(X)|=2$,

But how would I show this? I'm assuming I have to show something on the lines that $U_+$ and $U_-$ alone are both connected and path connected, however, when considered together i get that they're not connected and hence not path connected? therefore i have two components giving me that $|\pi_0(X)|=2$?

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Hints: The images of $f_+$ and $f_-$ are both closed, non-empty and disjoint in the topology of $X$. So think first about connectedness. Then recall what you know about continuous images of path connected sets.

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  • $\begingroup$ would i find clopen sets hence the topological space would not be connected? and hence not path connected? therefore it consists of two path components? $\endgroup$ – smith May 10 '15 at 22:33
  • $\begingroup$ @smith: Yes, it won't be connected. But the two components are path components. Note that $f_+$ and $f_-$ are continuous functions and the sets are continuous images of path connected sets. $\endgroup$ – T. Eskin May 10 '15 at 22:50

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