# A two-to-one continuous mapping

Does there exist a continuous map $f : [0, 1] \to [0, 1] \times [0, 1]$ such that the pre-image of any point of the square $[0, 1] \times [0, 1]$ contains precisely two points of the interval $[0, 1]$?

I guess the answer is no, but i have no idea how to consider this, even how to start? Anyone can help? Thanks!

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Is this homework? Please tag it if it is. –  Inquest Mar 14 '12 at 16:58
No. If the preimage of a point is $\{x_1,x_2\}$ with $x_1<x_2$, consider $U\subset[0,1]$ consisting of all such $x_1$. Similarly consider $V$ consisting of all $x_2$. Show that $U,V$ form a separation of $[0,1]$ contradicting that $[0,1]$ is connected.
Further hint: you also need to use that $[0,1]$ is compact.
I really have no reason to doubt this argument; just my own ability to verify it. Why are the sets $U,V$ open? –  Arthur Fischer Mar 14 '12 at 18:10