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Consider $X = T^3 \setminus \{p\}$, where $T^3 = S^1 \times S^1 \times S^1$ and $p$ is any point.
The cohomology ring $H^{\bullet}(X, \mathbb{Z})$ is the exterior algebra generated by three classes $u_1, u_2, u_3$ of degree 1, modulo the relation $u_1 u_2 u_3 = 0$. That is, it is $\Lambda_{\mathbb{Z}}[u_1, u_2, u_3] / (u_1 u_2 u_3)$, where the exterior algebra satisfies $u_i^2 = 0$ and $u_i u_j = -u_j u_i$ for $i \neq j$. This gives cohomology groups $H^0(X, \mathbb{Z}) \cong \mathbb{Z}$, $H^1(X, \mathbb{Z}) \cong \mathbb{Z}^3$, $H^2(X, \mathbb{Z}) \cong \mathbb{Z}^3$, and $H^k(X, \mathbb{Z}) = 0$ for $k \geq 3$.
Proof:
The space $T^3$ has a CW-complex structure with one 0-cell, three 1-cells, three 2-cells, and one 3-cell. The boundary maps in the cellular chain complex are all zero, so the homology groups are $H_0(T^3; \mathbb{Z}) \cong \mathbb{Z}$, $H_1(T^3; \mathbb{Z}) \cong \mathbb{Z}^3$, $H_2(T^3; \mathbb{Z}) \cong \mathbb{Z}^3$, $H_3(T^3; \mathbb{Z}) \cong \mathbb{Z}$. Since the homology groups are free, the cohomology groups are isomorphic: $H^k(T^3; \mathbb{Z}) \cong H_k(T^3; \mathbb{Z})$ for each $k$. The cohomology ring is the exterior algebra $\Lambda_{\mathbb{Z}}[u_1, u_2, u_3]$, where the $u_i$ are the dual basis to the 1-cycles generating $H_1$, with the relations $u_i^2 = 0$ and anticommutativity $u_i u_j = -u_j u_i$ for $i \neq j$, and the top class $u_1 u_2 u_3$ generates $H^3$.
Removing the point $p$, assume $p$ is in the interior of the 3-cell. Then $X$ is homotopy equivalent to the 2-skeleton of $T^3$, denoted $T^3_{(2)}$, via a deformation retraction: the 3-cell minus an interior point deformation retracts to its boundary $S^2$, fixing the boundary pointwise, and this retraction extends to the whole space because the boundary is already glued to the 2-skeleton.
The inclusion $i: T^3_{(2)} \hookrightarrow T^3$ induces isomorphisms on cohomology in degrees $\leq 2$, as the relative cohomology $H^k(T^3, T^3_{(2)}; \mathbb{Z}) \cong 0$ for $k \neq 3$ and $\mathbb{Z}$ for $k=3$. Thus, $H^k(X; \mathbb{Z}) \cong H^k(T^3; \mathbb{Z})$ for $k \leq 2$, and $H^3(X; \mathbb{Z}) = 0$.
The cup product structure on $H^{\leq 2}(X; \mathbb{Z})$ is induced from that on $T^3$ via $i^*$, so it coincides with the exterior algebra in degrees $\leq 2$. However, products landing in degree 3 are zero because $H^3(X; \mathbb{Z}) = 0$. This gives the stated ring.
Consider $Y = S^3 \setminus \{p_1, p_2, p_3\}$, with distinct points $p_1, p_2, p_3$.
The cohomology ring $H^{\bullet}(Y, \mathbb{Z})$ is $\mathbb{Z}[v_1, v_2]/(v_1^2, v_2^2, v_1 v_2)$, where each $v_i$ has degree 2. This gives cohomology groups $H^0(Y, \mathbb{Z}) \cong \mathbb{Z}$, $H^2(Y, \mathbb{Z}) \cong \mathbb{Z}^2$, and $H^k(Y, \mathbb{Z}) = 0$ for $k \neq 0,2$.
Proof:
By stereographic projection from $p_1$, $S^3 \setminus \{p_1\} \cong\mathbb{R}^3$, so $Y \cong\mathbb{R}^3 \setminus \{q_2, q_3\}$, where $q_2, q_3$ are the images of $p_2, p_3$. The space $\mathbb{R}^3 \setminus \{k$ points} is homotopy equivalent to the wedge of $(k-1)$ copies of $S^2$ for $k \geq 2$.
Thus, $Y \simeq S^2 \vee S^2$. The cohomology groups follow: $H^0(Y; \mathbb{Z}) \cong \mathbb{Z}$, $H^2(Y; \mathbb{Z}) \cong \mathbb{Z}^2$, and zero otherwise. For the ring, let $v_1, v_2$ be the generators of $H^2$ dual to the two $S^2$ factors. The square $v_i^2 = 0$ because each $S^2$ has zero cup square (as $H^4(S^2; \mathbb{Z}) = 0$). The cross product $v_1 v_2 = 0$ because the supports can be separated by disjoint neighborhoods in the wedge. |
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