The first problem is finding a map from S to T. The homotopy carries points in S along paths to their destinations in T, but these paths may merge and diverge. In its extreme, the homotopy could expand the single point S into a larger space T. That's hardly a homeomorphism; there isn't even a well defined function from S to T.
Let's make another assumption and see if that helps. Assume, for any x in S, there is a unique y in T, such that the homotopy carries x to y. Now there is a 1-1 function from S to T. Is it a homeomorphism?
We can't vector through D, because several points in D may map to x at time 0. As long as all these points map to y at time 1, we're ok. But along the way, these points may trace out different paths from x to y. Let's make another assumption. Assume there is a bijection from D to S and from D to T. This implies a bijection from S to T, and x cannot spawn a dozen paths that spin off in different directions and wind up at y. There is one path from x to y. Do we now have a homeomorphism from S to T?
The proof could get a little complicated if the various paths running from the points of S to the points of T collide. In the extreme, S could crunch down to a single point, then expand to T. Let's make another assumption. The function on D, throughout the homotopy, is always a bijection. Now the paths remain separate, all the way from S to T. Do we finally have a homeomorphism from S to T?
Let a space consist of two copies of the ray x ≤ 0, and the curves y = ±x2 for x > 0. The copies of the negative ray coincide. If -3a is in an open set, then so is -3b. Open sets are defined by the usual metric in the xy plane.
Let D be the space of two points, with the discrete topology. At time 0, map them onto -1a and -1b, which has the indiscrete topology. This is a continuous bijection. As time moves along, the images move linearly to 0a and 0b, then one image follows the upper parabola and one image follows the lower parabola, out to x = 1. Verify that this is a continuous map. The image at time 0 is connected, and the image at time 1 is not. The two images are not homeomorphic.
This may seem a bit contrived, because the space isn't even half hausdorff. Let's take a step back and assume the space is hausdorff, and D is compact, and the functions at time 0 and time 1 are bijections. A theorem is going to come to our rescue. A continuous bijection from a compact set onto a hausdorff space is a homeomorphism. Therefore S, D, and T are homeomorphic. The homotopy implies the homeomorphism.
Sometimes we make S the domain, so that S is "transformed" into T via the homotopy, and the resulting transformation implies S and T are homeomorphic. A homotopy stretches the sphere into an ellipse, hence the sphere and the ellipse are the same, topologically. Sometimes a continuous transformation is easier to visualize than the implied homeomorphism. But remember, a homotopy implies a homeomorphism only under certain conditions.
You may have heard topology described this way in an introductory text or lecture.
"If a sphere is made of infinitely flexible rubber, you cannot bend, twist, stretch, or otherwise deform the sphere to produce a torus. Topologically, they are distinct shapes."
This is true, but a bit misleading. Here is the real story.
There is no homotopy carrying the sphere onto the torus, because the sphere, which also acts as the domain D, is compact, and R3 is hausdorff, and a bijective homotopy between the two would imply a homeomorphism, yet the sphere and torus are not homeomorphic, as we showed in an earlier section.
Note that the lack of a transforming homotopy is almost a corollary. We used algebraic topology to prove the two shapes are distinct, hence there is no homotopy.
And the converse is not obvious either. There are plenty of homeomorphic spaces without connecting homotopies. Put two spheres in disjoint spaces, and they are obviously homeomorphic, even though there is no homotopy that carries one sphere onto the other.