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The complement of an embedded circle in $\mathbb{R}^3$, is homotopy equivalent to $S^1 \vee S^2$. To find its cohomology ring $H^*(X; \mathbb{Z})$ where $X = S^1 \vee S^2$, proceed as follows.
First, compute the cohomology groups. View $X$ as the wedge sum of $S^1$ and $S^2$, which has a CW structure with one 0-cell (the wedge point), one 1-cell (from $S^1$), and one 2-cell (from $S^2$). The cohomology groups are:
$H^0(X; \mathbb{Z}) \cong \mathbb{Z}$,
$H^1(X; \mathbb{Z}) \cong \mathbb{Z}$ (generated by the class $\alpha$ dual to the 1-cell),
$H^2(X; \mathbb{Z}) \cong \mathbb{Z}$ (generated by the class $\beta$ dual to the 2-cell),
$H^k(X; \mathbb{Z}) = 0$ for $k \geq 3$.
Next, determine the ring structure under the cup product.
$\alpha \cup \alpha \in H^2(X; \mathbb{Z})$: This corresponds to the square of the fundamental class of $S^1$, but $H^2(S^1; \mathbb{Z}) = 0$, and the attaching map ensures no nontrivial contribution from the wedge, so $\alpha^2 = 0$.
$\beta \cup \beta \in H^4(X; \mathbb{Z})$: This is zero since $H^4(X; \mathbb{Z}) = 0$.
$\alpha \cup \beta \in H^3(X; \mathbb{Z})$: The supports of $\alpha$ and $\beta$ intersect only at the wedge point (which does not contribute to the cup product), and $H^3(X; \mathbb{Z}) = 0$, so $\alpha \cup \beta = 0$.
$\beta \cup \alpha = (-1)^{1 \cdot 2} \alpha \cup \beta = \alpha \cup \beta = 0$.
Thus, the ring is generated by $\alpha$ and $\beta$ subject to these relations, yielding $H^*(X; \mathbb{Z}) \cong \mathbb{Z}[\alpha, \beta]/(\alpha^2, \beta^2, \alpha\beta)$ with $\deg \alpha = 1$ and $\deg \beta = 2$. |
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