Is $D^2\times S^1$ compact?

Question

I'm studying about the homology group and I have a question that denote $D^2$ the closed unit disk and $S^1$ the circle.The homology groups of a 3-dimension manifold $D^2\times S^1$ are $H_0(D^2\times S^1)=H_1(D^2\times S^1)\cong\Bbb Z$ and $H_n(D^2\times S^1)=0$ if $n= 2,3,...$ since $D^2\times S^1$ is homotopy equivalent to $S^1$.But I find a theorem form a Japanese text:

Let $M$ be a n-dimension compact connected manifold and then there is a nonnegative integer $l$ such that either $H_n(M)\cong\Bbb Z $ and $H_{n-1}(M)\cong\Bbb Z^l$ or $H_n(M)=0$ and $H_{n-1}(M)\cong(\Bbb Z/2\Bbb Z)\oplus \Bbb Z^l$.

This theorem is not valid for $D^2\times S^1$ but why? $D^2\times S^1$ is clearly connected so it is not compact? But it is closed and bounded in $\Bbb R^3$ so it is compact. Could you plaese tell me where did I make a mistake? Thank you!

Answer 1

This theorem is not valid for $D^2\times S^1$ but why? $D^2\times S^1$ is clearly connected so it is not compact?

It is compact. However $D^2\times S^1$ is not a manifold. At least not in the sense of the theorem. $D^2\times S^1$ is a manifold with (nonempty) boundary. But the theorem refers to manifolds without boundary, which is the usual understanding of the word "manifold" on its own.

This sounds like a small difference, but as the example above shows: not so small.