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For a singular $(k+l)$-simplex $\sigma :[v_0,\dots ,v_{k+l}]\to X$, define the cochain-level cup product byIf $\alpha \in C^k(X,A;R)$ and $\sigma$ lies in $A$, its front $k$-face also lies in $A$, so the first factor vanishes. Thus $\alpha ⌣\beta \in C^{k+l}(X,A;R)$.

For the projections from $X\times Y$, the exterior product in cohomology isFor CW pairs over a field $R$, subject to the usual finite-type condition that makes the graded tensor product commute with the relevant direct products, the Künneth theorem says thatMore generally the conclusion holds over a principal ideal domain when one factor has degreewise finitely generated free cohomology. It fails over $\mathbb{Z}$ in general: $H^{\ast }({\mathbb{RP}}^2;\mathbb{Z})$ is $\mathbb{Z}$ in degree zero and $\mathbb{Z}/2$ in degree two, but the integral Künneth and universal coefficient theorem for cohomology calculations giveThe tensor product of the two cohomology groups has no degree-three term, so $\Phi$ is not surjective.

Use the homeomorphismwhich carries the diagonal $\Delta$ to $T^2\times \{0\}$. HenceThe punctured torus deformation retracts onto a wedge of two circles. Let $a,b$ be the degree-one generators from the first torus and $c,d$ those from the punctured torus. Since all cohomology groups are free, the Künneth theorem identifies the integral cohomology ring asThus $ab$, $ac$, $ad$, $bc$, and $bd$ form a basis in degree two; $abc$ and $abd$ form a basis in degree three; and all higher positive degrees vanish.