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The degree-one Bockstein homomorphism makes modulo-$n$ cohomology into a cochain complex. Its cohomology is the Bockstein cohomology. For prime $n$ this is a graded vector space; for composite $n$ it is a graded $\mathbb{Z}/n$-module. For ${\mathbb{RP}}^3$ at modulus two, its only nonzero groups are ${\mathbb{F}}_2$ in degrees zero and three. This is the cohomological counterpart of Bockstein homology.

The cohomology of the cellular cochain complex naturally agrees with singular cohomology. A CW complex with no odd cells has zero cellular differentials and free integral cohomology on its even cells, although its cup product still requires a separate computation.

For a positively graded Lie algebra $L$ over $\mathbb{Q}$, finite-dimensional in each degree, its cochain algebra is free graded commutative on the graded dual of $sL$, with differential dual to the Lie bracket and its Koszul signs. The Jacobi identity gives $d^2=0$. A Lie element of degree $r$ contributes a cochain generator of degree $r+1$. Its cohomology computes ${\mathrm{Ext}}_{U(L)}(\mathbb{Q},\mathbb{Q})$ with total grading; the standard free enveloping-algebra resolution gives this identification. For a free graded Lie algebra, the tensor-algebra augmentation has a length-one free resolution, explaining the square-zero cohomology of a rational wedge sum of spheres.

For cochain maps $f,g:C^{\bullet }\to D^{\bullet }$, a cochain homotopy is a family $K:C^q\to D^{q-1}$ satisfyingCochain-homotopic maps induce the same map on cohomology.

A cochain map $f:E\to F$ consists of degreewise maps $f^i:E^i\to F^i$ satisfying $d_F^i{f}^i=f^{i+1}{d}_E^i$. This identity sends cocycles to cocycles and coboundaries to coboundaries, hence induces maps on cohomology. For complexes of sheaves the same identity holds as an identity of sheaf morphisms.

For a map between closed oriented manifolds of the same dimension, Poincare duality turns homological pushforward into a degree-preserving map on cohomology. It is adjoint to pullback under the duality pairing: $⟨f^{!}a⌣b,[N]⟩=⟨a⌣f^{\ast }b,[M]⟩$. If $f$ has degree $d$, nondegeneracy gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Maps of different dimensions require the corresponding degree shift.

An oriented normal bundle of a compact embedded submanifold has a Thom class. A tubular neighborhood and excision identify it with a relative class in $H^d(X,X\setminus Y)$. Its image in absolute cohomology is ${\epsilon }_Y$. It restricts to zero off Y and, in an oriented closed ambient manifold, represents the Poincare dual of Y with the normal-first induced orientation.

For the Hopf map $\eta$, the indicated four-cell attachment has cellular boundary multiplication by $f$ from dimension four to three. Its integral cohomology is $\mathbb{Z}$ in degrees zero and two, $\ker (f:\mathbb{Z}\to \mathbb{Z})$ in degree three, and $\mathbb{Z}/f\mathbb{Z}$ in degree four. If $x$ generates degree two and $z$ is the four-cell cochain class, then $x^2=dz$, with every other positive-degree product zero. Collapsing the three-sphere reduces the cup-square calculation to the Hopf invariant; the induced map on degree-four cohomology reduces its integer coefficient modulo $f$. When $f=0$ there is an extra free degree-three class, whose products still vanish by dimension.

For a cochain complex of sheaves $E$, its $i$th cohomology sheaf is ${\mathcal{H}}^i(E)=\ker (d^i)/\mathrm{im}(d^{i-1})$, where the image and quotient are taken in sheaves. Its stalk at $x$ is the cohomology of the stalk complex, because taking stalks is exact.

A short exact sequence of cochain complexes of sheaves $0\to E\to F\to G\to 0$ induces $\cdots \to {\mathcal{H}}^i(E)\to {\mathcal{H}}^i(F)\to {\mathcal{H}}^i(G)\to {\mathcal{H}}^{i+1}(E)\to \cdots$. Taking stalks is exact, so the construction and exactness follow from the ordinary long exact sequence of cohomology of complexes at every stalk. The stalkwise connecting maps come from the lift-and-differentiate construction and form a sheaf morphism.

Pull back along loop evaluation $S^1\times \Omega B\to B$ and take the slant product with the circle's fundamental class. This degree-lowering operation kills positive-degree cup products. It commutes with stable cohomology operations such as Steenrod squares and identifies a transgressing base class with its fiber class in the path-loop fibration.

A compact Lagrangian submanifold $L$ with vanishing first de Rham cohomology cannot be a fibre of a smooth positive-dimensional fiber bundle whose nearby fibres are also Lagrangian submanifolds. In a Weinstein neighborhood, nearby fibres are one-form graphs of closed forms. They are exact by the cohomology assumption, so meet the zero section at a critical point of their potential, contradicting disjointness of distinct fibres.

A compact smooth $n$-manifold with nonempty boundary embedded in $S^n$ is a proper compact locally contractible subset. Alexander duality therefore expresses the reduced homology of its complement by the reduced cohomology of the abstract manifold. Thus these groups do not depend on its particular smooth embedding.

For connected oriented closed $d$-dimensional manifolds with $d\ge 2$, their connected sum removes an open $d$-dimensional ball from each and identifies the resulting boundary spheres by an orientation-reversing diffeomorphism. The remaining orientations fit together. Collapsing the separating sphere gives a pinch map to the wedge sum of the two closed manifolds, and projection to either summand has degree of a continuous mapping one. In intermediate positive degrees the cohomology is the direct sum of the summand groups. Products of classes from different summands vanish; products landing in top degree use the single common orientation class. These statements follow from excision, the Mayer–Vietoris sequence and the degree-one projections.

The derived costalk at $x$ is $i_x^{!}E$, where $i_x:\{x\}\hookrightarrow X$. It computes local cohomology supported at $x$. On an oriented $N$-manifold the constant sheaf has costalk $\mathbb{Q}[-N]$, whose cohomology is in degree $N$.

Let $M,N$ be closed connected oriented manifolds of the same dimension, and suppose $a^r$ generates the top integral cohomology of $N$. If every class of degree $|a|$ on $M$ has zero rth cup power, every continuous map $f:M\to N$ has mapping degree zero. Indeed, $f^{\ast }(a^r)=(f^{\ast }a{)}^r=0$, and evaluation on the fundamental class gives $\mathrm{deg}f=0$. For $S^2\times S^4\to {\mathbb{CP}}^3$, the degree-two class is a multiple of the sphere class $u$, whose square vanishes.

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.

A degree of a map between oriented manifolds equal to one makes the pullback on degree-one cohomology injective by the Poincare duality pairing. Thus a degree-one map of closed oriented surfaces requires $2h\le 2g$. Conversely, collapse the extra handles in ${\Sigma }_h\#{\Sigma }_{g-h}$ onto a point. The retained oriented disc shows that this map has degree one.