Homology of $\mathbb{R}^3 - S^1$ I've been looking for a space on the internet for which I cannot write down the homology groups off the top of my head so I came across this:
Compute the homology of $X : = \mathbb{R}^3 - S^1$.
I thought that if I stretch $S^1$ to make it very large then it looks like a line, so $\mathbb{R}^3 - S^1 \simeq (\mathbb{R}^2 - (0,0)) \times \mathbb{R}$. Then squishing down this space and retracting it a bit will make it look like a circle, so $(\mathbb{R}^2 - (0,0)) \times \mathbb{R} \simeq S^1$. Then I compute 
$ H_0(X) = \mathbb{Z}$
$ H_1( X) = \mathbb{Z}$
$ H_n(X) = 0 (n > 1)$
Now I suspect something is wrong here because if you follow the link you will see that the OP computes $H_2(X,A) = \mathbb{Z}$. I'm not sure why he computes the relative homologies but if the space is "nice" then the relative homologies should be the same as the absolute ones, so I guess my reasoning above is flawed.
Maybe someone can point out to me what and then also explain to me when $H(X,A) = H(X)$. Thanks for your help!
Edit $\simeq$ here means homotopy equivalent.

Another way...
Consider $S^3$ as Alexandrov's compactification of $\mathbb{R}^3$: $S^3 = \mathbb{R}^3 \cup \{ \infty \}$. The set $X = \mathbb{R}^3 \setminus (S^1 \times \{ 0 \})$ can be seen as the complement in $S^3$ of the union $Y$ of a circle $S^1$ and of a point $P$. Since $S^3$ is homogeneous, we can suppose that $\infty$ is a point of $S^1$, so
$$
Y = S^3 \setminus (S^1 \cup \{ P \} ) = \mathbb{R}^3 \setminus ((S^1 \setminus \{ \text{a point} \}) \cup \{ P \}.
$$
Therefore $X$ is homeomorphic to the complement $Y$ in $\mathbb{R}^3$ of the union of a line and of a point: $Y = \mathbb{R}^3 \setminus (\{ x=y=0 \} \cup \{(1,0,0) \})$. Probably applying Mayer-Vietoris to $Y$ is easier than to $X$.
However it can be seen that $Y$ is homotopically equivalent to the very simple CW-complex $S^1 \vee S^2$, so computing homology is easy using cellular homology.
