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For a group action on $X$ and a contractible free $G$-space $EG$, the diagonal orbit space $EG{\times }_{G}X$ maps to the classifying space $BG$ with fibre $X$. It defines equivariant cohomology. When $X\to X/G$ is a principal $G$-bundle, the map $EG{\times }_{G}X\to X/G$ has contractible fibre $EG$ and gives a homotopy equivalence for spaces of CW type. Freeness without the bundle hypotheses should not be substituted for this assertion for arbitrary topological groups.

A cohomology operation is a natural transformation between cohomology functors. Thus for every continuous map $f$, its pullback commutes with the operation: $\theta (f^{\ast }a)=f^{\ast }\theta (a)$. This makes operations additional invariants of homotopy equivalence, beyond the additive groups and their cup products.

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.

If $A\subseteq X$ is contractible and $(X,A)$ has the homotopy extension property, the quotient mapis a homotopy equivalence. Extend a contraction of $A$ to $X$; its endpoint is constant on $A$ and therefore factors through $q$ to supply a homotopy inverse.

A deformation retraction of $X$ onto $A$ is a homotopy from ${\mathrm{id}}_X$ to a retraction $r:X\to A$ that keeps $A$ inside itself throughout. It makes the inclusion $A\hookrightarrow X$ a homotopy equivalence.

An integral homology equivalence between simply connected spaces is a weak homotopy equivalence. A first nonzero relative homotopy group would contradict the Relative Hurewicz theorem. Between CW complexes the ordinary Whitehead theorem then gives a homotopy equivalence.

The complement of a closed tubular neighborhood of a smooth plane conic has the displayed integral homology. The Excision theorem and Thom isomorphism theorem reduce the calculation to the relative long exact sequence: the ambient degree-four fundamental class restricts with coefficient one, while the degree-two map is intersection with the conic and has coefficient two. A collar neighborhood identifies the open and compact exterior homotopy equivalence types. The boundary instead has first homology $\mathbb{Z}/4$, from the Gysin sequence and the normal Euler class four.

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

A homotopy sphere is a closed manifold with the homotopy type of a sphere of the same dimension. In dimensions at least two, a closed simply connected integral homology sphere is a homotopy sphere: successive applications of the Hurewicz theorem produce a map from the sphere inducing an isomorphism on integral homology, and the homological Whitehead theorem makes it a homotopy equivalence. The topological generalized Poincare theorem identifies its topological type, while prescribed smooth structures require a separate question.

Two spaces have the same homotopy type when there is a homotopy equivalence between them. Their homology and homotopy groups are then isomorphic. In Morse theory, attaching a cell records a change in the homotopy type of a sublevel set.

Suppose two integral cohomology groups of a space are free of rank one. An integral isomorphism can send each chosen generator only to itself or its negative. After reduction modulo $p$, naturality of a Steenrod reduced power between those groups therefore permits its coefficient to change only by a sign, even though an abstract vector-space change of basis could rescale by any nonzero field element. At $p=5$, coefficients one and two cannot be related by signs, so they obstruct an integral homotopy equivalence. This constraint is stronger than comparing the isolated mod-five Steenrod modules with arbitrary bases.

For a nonempty compact topological manifold $X^d$ without boundary, its suspension is a topological manifold without boundary exactly when $X$ is homeomorphic to $S^d$. The forward implication is a local homology from a link calculation at a suspension vertex: ${\tilde{H}}_j(X)$ is $\mathbb{Z}$ in degree $d$ and zero elsewhere. For $d\ge 2$, put a coordinate ball between two nested cone neighbourhoods. The inclusion of their punctures is a homotopy equivalence but factors through the simply connected punctured ball, forcing ${\pi }_1(X)=0$. Thus $X$ is a homotopy sphere, and the topological generalized Poincare theorem gives the result. The dimensions zero and one follow from the finite-set and circle classifications. Conversely $S(S^d)\cong S^{d+1}$. If manifolds with boundary are allowed, the interval and its disk suspension show why this formulation needs the boundary restriction.

For a map $f:X\to B$, let $E_f$ consist of points of $X$ and paths starting at their images, with projection $p(x,\gamma )=\gamma (1)$. Appending segments of a base homotopy proves the homotopy lifting property, so $p$ is a Serre fibration. The constant-path inclusion $s:X\to E_f$ satisfies $ps=f$ and is a homotopy equivalence: forget the path for an inverse, and shorten paths to prove the homotopy. The homotopy need not preserve the endpoint in $B$, so the statement concerns an equivalence of total spaces over a commuting triangle, not a fibrewise inverse. The fibre over $b_0$ is the homotopy fibre of $f$.