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I see that the Peano curve is a continuous surjection from the unit interval to the unit square (correct me if I'm wrong). Does it then follow that there is a continuous injection from the unit square to the unit interval?

Thank you!

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up vote 6 down vote accepted

The other three answers provide reasons for why there is no continuous injection from the unit square to the unit interval. But I wanted to show why you're specific argument fails.

The issue is that a continuous surjection $f:[0,1]\rightarrow [0,1]^2 $ will fail to be injective. The quick reason is that an injective continuous map between compact Hausdorff spaces is automatcally a homeomorphism onto its image, and $[0,1]$ and $[0,1]^2$ are not homeomorphic (as the other answers show).

So, since $f$ is surjective, there is an inverse injective function $g:[0,1]^2\rightarrow [0,1]$, but it involves making many choices. This is because whenever you have $y=f(x_1) = f(x_2)$ with $x_1\neq x_2$, then you must make a choice for $g(y)$. Should $g(y) = x_1$ or $g(y) = x_2$? (In worse cases, there aren't just $2$ different $x$s mapping to the same $y$, but sometimes infinitely many).

How do you make such a choice? Well you can, but not in any canonical fashion. Hence, while you do get an injective function $g:[0,1]^2\rightarrow [0,1]$, due to all the choices you had to make, it won't be continuous.

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No. The image of the square under a continuous function to the real line is a compact interval. Removing the center of the square leaves it connected. Removing an interior point from an interval disconnects it.

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Suppose that $f:[0,1]^2\to[0,1]$ is a continuous injection. $[0,1]^2$ is compact and connected, so $f\big[[0,1]^2\big]$ must be compact and connected; since it’s not a singleton, it must be a closed interval, and without loss of generality we may assume that it is $[0,1]$ itself, so that $f$ is a continuous bijection.

Now observe that $[0,1]\setminus\left\{\frac12\right\}$ is not connected, but $[0,1]^2\setminus f^{-1}\left(\frac12\right)$ is connected, and therefore $f$ is not continuous after all.

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No. Invariance of domain implies that this is not possible.

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