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Consider the map $\varphi : \mathbb R^2 \to \mathbb R^2$ defined by $(x,y) \mapsto (x,y^2)$.

Apparently this map is called a fold as the $(x,y)$-plane is folded over and creased along the axis $y=0$.

But I really don't see how this is the case: $y^2$ does not "fold" or "crease" anything it just bends the plane very very slightly. No?

Please could someone explain to me how this map folds the plane? I obviously misunderstand it completely.

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Consider the one-dimensional analogue: $\varphi : \mathbb{R} \to \mathbb{R}$, $y \mapsto y^2$. The image of $\varphi$ is $[0, \infty)$; furthermore, for all $x \in (0, \infty)$, $\varphi^{-1}(x) = \{\sqrt{x}, -\sqrt{x}\}$, and $\varphi^{-1}(0) = \{0\}$. In a sense, we have 'folded' $(-\infty, 0)$ onto $(0, \infty)$.

Added Later: The map $y \mapsto |y|$ would be a much better candidate for a 'fold'. For example, under this map $-2$ gets 'folded' onto $2$, whereas in the case above, $-2$ and $2$ both get mapped to $4$. This is more like a fold followed by a stretching of $[0, \infty)$. Having said that, the map $y \mapsto y^2$ will often be more favourable than the map $y \mapsto |y|$ as the former is smooth, while the latter is not differentiable at $0$.

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The map is 2:1 away from the $x$-axis, so both positive and negative $y$-values get mapped to positive $y$-values. The entire bottom half plane gets mapped to the upper half plane.

If it helps, think of this as a composition: $(x,y)\mapsto (x,|y|)\mapsto (x,|y|^2)$. It's easier to see the absolute value map as a crease, I think.

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    $\begingroup$ The map is not 2:1, $(0,0)$ has only one preimage. $\endgroup$ – Najib Idrissi May 29 '15 at 9:58
  • $\begingroup$ Tsk, tsk. Thanks. $\endgroup$ – Neal May 29 '15 at 11:08
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It's a fold because every point $(x,y)$ has another point $(x,-y)$ that also maps to the same point in the image $(x, y^2)$.

The domain is split by a crease (the y-axis) into two half planes which then both map one-to-one and onto the same image.   So it is a two-to-one mapping.   That's the folding.

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