Can you describe what is $S^1 \times$ [0,$\infty)$? Where "$\times$" stands for the product topology and the two factors are with the euclidean topology.

EDIT: What about $S^1 \times S^1$?

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    $\begingroup$ It's a half-infinite cylinder. $\endgroup$ – Parcly Taxel Jan 30 '18 at 16:32
  • 1
    $\begingroup$ Think of an infinite tube or an infinitely long hose. Now cut it in half (i.e., anywhere). $\endgroup$ – Michael Burr Jan 30 '18 at 16:34
  • $\begingroup$ What exactly are you looking for that the description "S^1 \times [0, \infty)$ doesn't already give you? $\endgroup$ – anomaly Jan 30 '18 at 16:35
  • $\begingroup$ It is the same as the closed unit disc minus the origin. $\endgroup$ – orole Jan 30 '18 at 16:39
  • $\begingroup$ @ParclyTaxel What about $S^1 \times S^1$? $\endgroup$ – qcc101 Jan 30 '18 at 16:47

One way to think of the cross product, is that at each point in the first factor, you are attaching a copy of the second space.

For example: $S^1 \times \{s\}$, where $s$ is a single point, is really just $S^1$, since at each point you are replacing it with a different point.

$S^1 \times \{s,t\}$ is a two copies of the circle. You can view this by taking a circle, and at each point, you are replacing it with two points, and looking at the full collection of these gives two circles.

Actually, $S^1 \times \{1, \dots, n\}$ is nothing but $n$ circles, and $S^1 \times \mathbb Z$ is a countable collection of circles. You can visualize them as stacked along some verticle axis, with a circle at each integer.

Going further, $S^1 \times \mathbb R$ is a circle, but whenever there was a point, you replace it with a line, so you get a circle of lines, or in other words, a cylinder.

$S^1 \times [a,\infty)$ is the same, but with a half open interval.

$S^1 \times S^1$ is a circle of circles, so at each point you attach a circle (for the sake of visualization, say you attach a circle with smaller radius), then you get a torus, with the traditional donut visual.

  • $\begingroup$ Addendum, this point of view lends itself well to one generalization of the cross product, known as a fiber bundle. Maybe just an idea to be aware of. One such example would be a mobius band, which looks kind of like $S^1 \times [0,1]$, but with a twist. $\endgroup$ – Andres Mejia Jan 30 '18 at 17:01

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