Can either pair of opposite sides of an arbitrary parallelogram be brought into coincidence isometrically in 3-space?

Question

Let P denote any nondegenerate planar parallogram, and let A and B be either pair of its opposite edges.

Does there always exist a continuous family of locally-isometric mappings h_t of P into 3-space, parametrized by the unit interval [0,1], such that the mapping h_t is an isometric embedding for t ∈ [0, 1), and h₁ is one-to-one except for mapping A and B each isometrically onto the same line segment in 3-space (with the same orientation)?

(This would mean that the image h₁(P) is topologically the cylinder S¹ × [0, 1].)

I am calling a mapping "isometric" when the metric on the domain equals the intrinsic metric along its image.

(I ask because I was quite surprised to find that, with a paper model, this appears to be eminently possible for a 60º-120º rhombus.)

Answer 1

Although I was able to solve this, I then realized that a correspondent had already explained the gist of the answer to me earlier, in a way I had not understood at first.

Take the parallelogram P and tile the plane with it by using two of its sides to generate a lattice of translations, as usual.

Now suppose we have a bi-infinite right circular cylinder Cyl whose circumference equals the length of one of the sides S of one instance of P (that we'll call P) in the plane. Wrap this tessellated plane around Cyl so that S becomes exactly one geodesic circumference of Cyl. Then it follows immediately that the side S' of P that is opposite to S must also become another circumference of the cylinder.

This implies that the sides T and T' of P that are not S or S' must exactly meet along their entire length.