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Can a hemispherical surface be flattened if tearing is allowed but no stretching is allowed in the interiors ?

A flattening, in this case, would be a continuous mapping from the hemispherical surface to a subset of $\mathbb{R}^2$ such that the arc-length distance between any two hemispherical points is equal to the Euclidean distance between their forward images. This does not exist.

I am considering a less stringent notion in the hopes that one may exist. I partition the hemispherical surface into finitely many pieces, (their boundary form the seams). The forward images of each connected piece are connected, mutually non-overlapping and metric preserving.

Using appropriately shaped gores or lunes constructed from flat unstretchable paper it is possible to make a hemispherical globe approximately.

For example, Shape of a flattened wedge from the surface of a sphere (3Blue1Brown video). (Link courtesy @sirous)

Related: Deriving surface area of a sphere using triangles

Related: Deriving the formula of the surface of a sphere using triangles.

I am interested in the theoretical zero-distortion case. Is this even possible ?

Tearing at the seams on a hemisphere are ok (like in a tennis ball) but no stretching is allowed in the interiors (unlike in the tennis ball case).

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$ Commented yesterday

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Comment: I think this can be a way.

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    $\begingroup$ No. You can't flatten the pieces without distorting either the width profile or the shape of the boundary, so you would need stretching. $\endgroup$ Commented yesterday
  • $\begingroup$ I upvoted for the attempt. Thanks for the try. Let me show an area where there is stretching in this example. If the horizontal line through the center maps to the equator then the central vertical line of the lobe and its two curved edges ought to be the longitudes (perp to equator). These longitudes analogues are different length, so stretching is involved. One can take out the excess length by cutting out a bow-tie shaped piece from center of the lens like shape. Hard to judge it suffices without details about the curves. Are these sines $\endgroup$ Commented yesterday

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