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Attaching a $q$-cell along $\varphi :S^{q-1}\to X$ forms the quotient of $X\bigsqcup D^q$ identifying each boundary point with its image under $\varphi$. A constant attaching map gives $X\vee S^q$. An attaching map can affect the cohomology ring even when it does not change the additive cohomology groups.

A degree-$q$ cohomology group is the quotient of degree-$q$ cocycles by coboundaries in a cochain complex. Topological cohomology groups arise from singular or cellular cochains. Their direct sum carries the cohomology ring structure when equipped with the cup product.

For positive unequal dimensions $p,q$, let $u,v$ be the pullbacks of the integral orientation classes of the factors. The Künneth theorem gives the free abelian basis $1,u,v,uv$, with degrees $0,p,q,p+q$. The products satisfy $u^2=v^2=0$ and $vu=(-1{)}^{pq}uv$, by naturality and graded commutativity of the cup product. In particular, for $p=2,q=4$, the cohomology ring is $\mathbb{Z}[u,v]/(u^2,v^2)$, with degrees two and four.

Since $T^r$ is a product of circles, the Künneth theorem identifies its cohomology ring with the exterior algebra on the coordinate degree-one classes. Inversion negates each integral degree-one class, so it acts as $(-1{)}^q$ in degree $q$. Even in characteristic two, each degree-one class has square zero.

The cell structure gives one integral cohomology generator in each degree divisible by four through $4n$. The Gysin sequence of a sphere bundle for the quaternionic tautological line bundle makes multiplication by its Euler class an isomorphism between successive such degrees. Choose $u$ to be the negative of that Euler class; its powers generate the displayed cohomology ring.

The antipodal map on the two-sphere reverses orientation, so its mapping torus is a nonorientable closed three-manifold. The Wang sequence gives $H_0=H_1=\mathbb{Z}$, $H_2=\mathbb{Z}/2$ and $H_3=0$. The universal coefficient theorem for cohomology therefore gives the cohomology ring $\mathbb{Z}[t,u]/(t^2,tu,u^2,2u)$ with $|t|=1$, $|u|=3$. The base-circle class satisfies $t^2=0$ by pullback; the other positive-degree products vanish for dimension reasons.

For $k\ge 1$, put $U={\mathbb{CP}}^{2k}\setminus {\mathbb{RP}}^{2k}$. Its cohomology ring isHere $x$ is the restricted hyperplane class. The cohomological Gysin map of an embedding sends $a^{2j}$ to $h^{k+j}$ and every odd-degree class to zero, by Poincare duality and mod-two restriction from complex to real projective space. Its exact sequence gives one-dimensional groups in even degrees $0,2,\dots ,4k-2$, and zero otherwise. Choose $y$ with $\delta y=a$. The connecting-map module identity gives $\delta (x^{j}y)=a^{2j+1}$ for $0\le j<k$, so these are the upper-half generators. The lower-half generators are $x^j$. The relations $x^k=0$ and $y^2=0$ follow respectively from $i_{!}1=h^k$ and $H^{4k}(U;{\mathbb{F}}_2)=0$. The result is a ring isomorphism with Complex projective space ${\mathbb{CP}}^{k-1}$ times $S^{2k}$; it does not by itself assert a homotopy equivalence of the spaces.

For a cohomology class $a$, its rth cup power is its r-fold product in the cohomology ring. Naturality gives $f^{\ast }(a^r)=(f^{\ast }a{)}^r$. A nilpotent class has some positive cup power zero; the nilpotence order is preserved by ring isomorphisms and can obstruct a homotopy equivalence or a nonzero mapping degree.

For even $n\ge 2$, attach an $(n+1)$-cell to $S^n\times S^n$ along the diagonal. The cellular chain complex has boundary $(1,1)$ for the new cell. The degree-$n$ cohomology generator restricts to $a-b$, and the top generator restricts to $ab$. Naturality of the cup product therefore gives $u^2=-2v$, since $(a-b{)}^2=-2ab$. Thus the resulting cohomology ring is $\mathbb{Z}[u,v]/(u^2+2v,uv,v^2)$, with $|u|=n$ and $|v|=2n$. This distinguishes the attachment from a wedge of spheres, whose positive-degree products vanish.

Give $(S^2{)}^m$ its product orientation and ${\mathbb{CP}}^m$ its complex orientation. If $u_i$ are the degree-two sphere classes and $h$ the normalized projective class, write $f^{\ast }h={\sum }_i{a}_i{u}_i$. In the source cohomology ring, $u_i^2=0$ and the generators commute. Hence $f^{\ast }{h}^m=m!({\prod }_i{a}_i)u_1\cdots u_m$. Evaluation on the fundamental class gives the displayed mapping degree. Every positive degree is therefore at least $m!$, and the polynomial multiplication map to complex projective space attains that bound.

Homotopic continuous maps induce the same map on cohomology. In particular, a homotopy equivalence induces an isomorphism of cohomology rings.

Integral cohomology is cohomology with integer coefficients, $H^{\ast }(X;\mathbb{Z})$. Its cup product makes it a graded cohomology ring, retaining torsion and integral information that can disappear after changing coefficients to a field.

Use the complement bundle on a complex Stiefel manifold inductively. If the ring for $V_k$ is exterior with smallest positive degree $2n-2k+1$, the complement bundle has Euler degree $2(n-k)$, in which base cohomology vanishes. Its Euler class is consequently zero. The Gysin ring splitting for an odd-dimensional sphere bundle adds one exterior generator of degree $2(n-k)-1$. The initial base $V_0$ is a point. This gives the full integral cohomology ring and in particular its torsion-freeness. At $k=n$, the unitary group has Poincare polynomial ${\prod }_{j=0}^{n-1}(1+t^{2j+1})$.

For $m\ge 0$, the cohomology ring isandThe generator $a$ reduces to the square of the degree-one generator in the mod-two cohomology ring of real projective space; $b$ is the integral orientation class in odd top dimension.

The wedge ${\mathbb{RP}}^2\vee S^1$ has the same additive integral cohomology as the Klein bottle: $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in degree one, and $\mathbb{Z}/2$ in degree two. Products of positive-degree classes vanish because the degree-one class comes from the circle summand, so its cohomology ring is isomorphic to that of the Klein bottle.

Two finite CW complexes can have isomorphic integral cohomology rings but different cohomology rings modulo $p$. For example, attaching a three-cell of degree $p$ to the projective line in ${\mathbb{CP}}^2$ preserves the nonzero square of the degree-two class modulo $p$, whereas the corresponding Moore space wedged with $S^4$ has zero square. The spaces therefore cannot be homotopy equivalent.

Modulo two the antipodal map acts trivially on fiber cohomology. The Wang sequence supplies a class $y$ restricting to the generator of $H^2(S^2;{\mathbb{F}}_2)$. The Leray-Hirsch theorem with classes $1,y$ gives a basis $1,t,y,ty$, so $ty\ne 0$. Pullback from the circle gives $t^2=0$, while dimension gives $y^2=0$. This cohomology ring agrees with that of $S^1\times S^2$, although the original manifold is nonorientable.

The cohomology ring of the product of two real projective planes is ${\mathbb{F}}_2[x,y]/(x^3,y^3)$, with $|x|=|y|=1$. Pairing against $x^2$, $xy$ and $y^2$ identifies the Poincare dual of the diagonal as $x^2+xy+y^2$. For a self-map $f$ with $f^{\ast }x=\epsilon x$, the graph pullback evaluates on the mod-two fundamental class as $1+\epsilon +{\epsilon }^2=1$. Thus no graph of a map on the whole projective plane can avoid the diagonal. The argument uses a relative Thom class and does not assume transverse intersections.

The standard CW complex structure on Complex projective space ${\mathbb{CP}}^3$ has one cell in each dimension $0,2,4,6$ and no other cells. Its cellular chain complex therefore has zero differentials: each boundary has an odd-dimensional target group, which is zero. The product of the usual two-cell structures on $S^2$ and $S^4$ likewise has exactly one cell in dimensions $0,2,4,6$. Thus cellular homology gives, for both spaces,The cohomology rings detect more than these additive groups. Let $c\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ be the hyperplane class. The cohomology ring of complex projective space isFor the normalization, a hyperplane is Poincare dual to $c$; intersecting three generic hyperplanes gives one point with positive complex orientation, so $⟨c^3,[{\mathbb{CP}}^3]⟩=1$. Together with the computed additive groups this makes $c,c^2,c^3$ generators in their respective degrees.

Let $u$ and $v$ be the pullbacks of the orientation classes of $S^2$ and $S^4$. The Künneth theorem and naturality of the cup product givewhere $uv$ generates degree six. In particular, every degree-two class on the product has square zero, whereas $c^2\ne 0$. Any graded ring isomorphism would send $c$ to $\pm u$, contradicting this product relation. A homotopy equivalence induces an isomorphism of cohomology rings, so the spaces have isomorphic integral homology but are not homotopy equivalent.