Homology of product

Question

I am suppsosed to compute singular homology of $\mathbb{S}^1\times\mathbb{D}^2-\{(1,0)\}$, namely $H_*(X;\mathbb Z)$ for my space $X$, but I don't even have an idea of which tools to use.

Can anyone help?

Answer 1

Removing the point $(1,0)$ from $S^1\times D^2$ doesn't make much sense to me unless further explanation is provided by the OP in terms of notation. I assume that the space in question is: $$ S^1\times (D^2\setminus\{(1,0)\}) $$ even though in this case $D^2$ is the open disc and $(1,0)$ doesn't belong in $D^2$!! So it might be that $D^2$ denotes the closed 2-disk. The OP doesn't clarify.

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In any case $D^2\setminus\{(1,0)\}$ is contractible hence $S^1\times (D^2\setminus\{(1,0)\}) $ has the same homology groups as $S^1\times \{pt\}$ i.e. the same homology groups as $S^1$.

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$H_0=\mathbb{Z}$ since $S^1$ is path-connected and $H_1=\mathbb{Z}$ since $\pi_1(S^1)=\mathbb{Z}$ is abelian. $H_k=0$ for $k\geq 2$ since $S^1$ is a one-dimensional manifold.

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If the OP meant that from $S^1\times D^2$ we remove $\{(p,q)\}$ with $p\in S^1$ and $q\in D^2$ then $S^1\times D^2 \setminus \{(p,q)\}$ is the interior of the 2-torus with a point removed, which is contractible i.e. has the homology of a point.

Answer 2

If the intention was $S^1 \times (D^2 - \{(1,0)\})$, then the following works:

$D^2 - \{1,0\}$ is contractible, so it is enough to consider $S^1$. In this case, one $H_1= \mathbb Z$. One can think about it as just "Fattening up $S^1$ (minus a line). $H_n$ is trivial for all $n>1$.

Let $f:(D^2-\{(1,0)\}) \times [0,1]$ be defined by $\mathbf{x} \mapsto (1-t) \mathbf{x}$ for all $\mathbf{x},t \in (D^2-\{(1,0)\}) \times [0,1]$. This defines a retraction onto $\mathbf{0}$.

Note that it matters that the point missing is $(1,0)$, since otherwise there would be an "obstruction" in the straight line contraction to the origin.