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For maps $A\to X$ and $A\to Y$, the double mapping cylinder $X\bigsqcup (A\times I)\bigsqcup Y$, with each endpoint attached by its given map, models the homotopy pushout. A map out of this model is a pair of maps on $X$ and $Y$ together with a homotopy between their restrictions to $A$. This is why compatible homotopy classes extend over a homotopy pushout. For a CW pair, the inclusion is a cofibration and the ordinary pushout gives the same homotopy type.

A map $p:E\to B$ is a Serre fibration if every homotopy $H:D^k\times I\to B$, $k\ge 0$, with a lift of its initial slice has a lift on the whole cylinder extending that initial lift. By attaching cells, this gives the relative homotopy lifting property for CW pairs: a specified lift on the initial slice and on the subcomplex cylinder can be extended.

Give $PY$ the compact-open topology. Suppose the initial lifting is a family of paths ${\gamma }_x$ starting at $y_0$, with ${\gamma }_x(1)=H(x,0)$. DefineThe two expressions coincide on their common boundary, the initial path at $s=0$ is exactly ${\gamma }_x$, and the endpoint is $H(x,s)$. The adjoint map on $D^k\times I\times I$ is continuous by pasting; compactness of the path parameter gives continuity into the path space. Thus endpoint evaluation $PY\to Y$ is a Serre fibration. The variable subdivision avoids changing the initial lift by an unwanted reparametrization.

For a based pair $(E,F,e_0)$, the long exact sequence of relative homotopy groups isending in ${\pi }_1(E)\to {\pi }_1(E,F)\to {\pi }_0(F)\to {\pi }_0(E)$. The low-dimensional end is an exact sequence of pointed sets with its usual actions, rather than an entirely abelian sequence.

The boundary sends a relative disk map to its restriction on the boundary sphere. An absolute class has constant boundary, so $\partial j_{\ast }=0$. Conversely, if a relative map has null-homotopic boundary in $F$, choose a based null-homotopy there. Attach this null-homotopy as a collar to the disk, leaving the original map on a smaller inner disk. The resulting boundary is constant at $e_0$, so it defines an absolute class in ${\pi }_n(E)$; inserting the collar is a relative homotopy, so its image is the original relative class. This proves $\mathrm{im}{j}_{\ast }=\ker \partial$. At $n=1$, append a path in $F$ from the terminal point to $e_0$, giving the corresponding based loop and pointed-set exactness.

Now suppose $F=p^{-1}(b_0)$. Use a cube model in which the bottom and side faces are based and the top face maps into $F$. A based $n$-cube in $B$ lifts from its constant bottom face, with sides held at $e_0$ by relative lifting. Its top face lies in $F$, proving surjectivity of the relative homotopy of a Serre fibration comparison $p_{\ast }:{\pi }_n(E,F)\to {\pi }_n(B,b_0)$.

For injectivity, lift a based homotopy of two projected cubes from the first relative lift; this ends at some lift of the second base cube. Any two lifts of that same base cube are relatively homotopic: insert an extra parameter between them, prescribe the two liftings on its two ends and the constant lift on the bottom, and apply relative lifting in the remaining cube direction. The top face stays in $F$ because its projection is $b_0$. Combining these homotopies proves injectivity. ThereforeAt $n=1$ the same construction gives a bijection of pointed sets, and one can transport the base fundamental-group law across it. An arbitrary pair's relative ${\pi }_1$ has no general group law. The printed target index $a$ is a typographical error: the map preserves degree $n$.

The last-column map $U(n)\to S^{2n-1}$ is surjective because a unit vector can be completed to a unitary basis. Its stabilizer is the block subgroup $\mathrm{diag}(U(n-1),1)$, so it induces a continuous bijection $U(n)/U(n-1)\to S^{2n-1}$. Unitary matrices form a compact set in their matrix-space topology; the quotient is compact and the sphere Hausdorff. HenceThis quotient is a fiber bundle. Near the last standard unit vector, project the other standard vectors by $I-v{v}^{\ast }$ onto $v^{\perp }$ and apply Gram-Schmidt orthogonalization; their independence persists in a small neighborhood. The resulting orthonormal columns together with $v$ give a continuous local section. Translating these sections covers the sphere and trivializes the projection, so it is a Serre fibration.

Apply this to $U(n)\to U(n+1)\to S^{2n+1}$. Its homotopy sequence containsFor $i\le 2n-1$, both outer groups vanish by cellular approximation theorem. Thus ${\pi }_i(U(n))\cong {\pi }_i(U(n+1))$ in the stated range. The component assertion also holds: diagonalizing a unitary matrix and varying its eigenvalue phases continuously connects it to the identity. This establishes the unitary sphere quotient and stability range.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

Let $C_{\ast }(X)$ be the integral singular chain complex. Augment it by the homomorphism $C_0(X)\to \mathbb{Z}$ sending every point simplex to one. The homology of this augmented complex is reduced homology: for a nonempty space, it agrees with ordinary homology in positive degrees and ${\tilde{H}}_0(X)$ is the kernel of $H_0(X)\to \mathbb{Z}$. The relative homology $H_{\ast }(X,A)$ is the homology of $C_{\ast }(X)/C_{\ast }(A)$; a relative cycle has boundary in $A$, and chains differing by a boundary or a chain in $A$ represent the same class.

Excision theorem: if $U\subset A\subset X$ and $\overline{U}\subset {\mathrm{int}}_{X}A$, the inclusion $(X\setminus U,A\setminus U)\hookrightarrow (X,A)$ induces isomorphisms on all relative homology groups. In particular, for a nonempty CW subcomplex $A$, choose a neighborhood $N$ deforming onto $A$, with the deformation fixing $A$. Such neighborhoods exist for CW pairs. Then $N/A$ is contractible. The long exact sequence of a pair gives $H_{\ast }(X,A)\cong H_{\ast }(X,N)$ and $H_{\ast }(X/A,\{\ast \})\cong H_{\ast }(X/A,N/A)$. Excision removes $A$ and the collapsed point respectively; the resulting pairs $(X\setminus A,N\setminus A)$ and $((X/A)\setminus \{\ast \},(N/A)\setminus \{\ast \})$ are homeomorphic. ConsequentlyHere the nonempty-subcomplex condition is the usual quotient convention in this statement. If $A=\varnothing$, use the pointed cofiber $X_{+}$ with a disjoint basepoint; the literal unpointed quotient $X/\varnothing =X$ would not give the asserted degree-zero identity.

For one embedded circle, its long exact sequence of a pair reduces toA separating circle bounds a subsurface, so its homology class is zero. A nonseparating circle has a transverse closed curve meeting it once, showing that its class is nonzero and primitive. Thus $i_{\ast }$ is either zero or an injection with torsion-free cokernel. The resulting possibilities arewith all other relative groups zero. These ranks determine whether the circle is null-homologous.

For $g\ge 2$, a disk boundary and a separating circle bounding a once-punctured torus have the same relative groups. The first is null-homotopic; the second is not. For a direct verification, use the surface-group presentation $⟨a_1,b_1,\dots ,a_g,b_g\mid {\prod }_i[a_i,b_i]=1⟩$. Send $a_1,b_1$ to free generators $u,v$, send $a_2,b_2$ to $v,u$, and all remaining generators to one. The relation maps to $[u,v][v,u]=1$, so this defines a homomorphism to the free group. The separating circle represents $[a_1,b_1]$ and maps to the nonempty reduced word $uv{u}^{-1}{v}^{-1}$; it is therefore not null-homotopic. This proves the requested failure to detect nullhomotopy. There is a necessary low-genus qualification: on ${\Sigma }_0$ or ${\Sigma }_1$, an embedded circle is nullhomologous exactly when it is nullhomotopic. On the sphere every circle bounds a disk; on the torus every essential embedded circle is nonseparating. The printed nondetection assertion therefore needs $g\ge 2$.

For two disjoint embedded circles let $r$ be the rank of their image in $H_1({\Sigma }_g)$, and let $c$ be the number of connected components of the cut surface. Removing open annular neighborhoods and applying excision gives the compact cut surface $W$ relative to its boundary. Poincare-Lefschetz duality gives $H_2(W,\partial W)\cong H^0(W)={\mathbb{Z}}^c$ and $H_1(W,\partial W)\cong H^1(W)$, which is free. Thus there is no hidden torsion. The long exact sequence gives $c=3-r$, and the Euler characteristic of the pair is $2-2g$, givingThe possibilities are $r=0,1,2$: two separating circles give zero; a nonseparating circle with a disk boundary, or two parallel essential circles, give one; circles around two different handles give two. For genus zero only $r=0$ occurs; for genus one $r=0,1$ occur; for every genus at least two all three occur. This is relative homology of a surface modulo disjoint circles.

The exactness axiom of a generalized cohomology theory gives a natural long exact sequence for each CW pair:For $(X,X)$, the restriction $h^n(X)\to h^n(X)$ is the identity. Its preceding relative-to-absolute map has zero image. The previous connecting map is zero too, since it follows the surjective identity in degree $n-1$. Exactness at the relative group therefore makes its map both injective and zero, givingin particular $h^n(x_0,x_0)=0$. No dimension axiom is being used; a generalized theory can have nonzero $h^n(x_0)$ in many degrees.

The wedge-product notation for pairs here means the smash product of CW pairs,Its quotient is the based smash product $(X/A)\wedge (Y/B)$. A multiplicative theory has an external productIf the two spaces are the same, pullback by the diagonal gives the relative cup product in $h^{k+m}(X,A\cup B)$. This is different from the wedge sum of two based spaces.

First read the covering subcomplexes as pointed subcomplexes, so $x_0\in A_j$. Contractibility makes the inclusion $\{x_0\}\hookrightarrow A_j$ a homotopy equivalence, and thus $h^r(A_j,x_0)=0$ in every degree, by exactness and homotopy invariance of cohomology. The relative exact sequence for the triple $\{x_0\}\subseteq A_j\subseteq X$ givesConsequently each homogeneous $w_j$ lifts to $v_j\in h^{r_j}(X,A_j)$. Repeated relative multiplication givesNaturality maps this product to the product of the $w_j$ in $h^{\ast }(X,x_0)$. ThereforeFor nonhomogeneous classes the same conclusion follows by distributivity. This is nilpotence from a contractible subcomplex cover.

There is an alternative sufficient convention: if $X$ is path connected, the $A_j$ need not contain $x_0$. The absolute image $\bar{w}$ of a reduced class restricts to zero at $x_0$, hence at any other point since the point inclusions are homotopic along a path. On a nonempty contractible $A_j$ it therefore restricts to zero, so exactness lifts ${\bar{w}}_j$ to $h^{\ast }(X,A_j)$. Their relative product vanishes as above. The map $h^{\ast }(X,x_0)\to h^{\ast }(X)$ is injective: restriction to the basepoint is split surjective by the map $X\to \mathrm{pt}$, making the previous connecting map zero. Thus vanishing of the absolute product gives vanishing of the reduced product too.

Without either the pointed-subcomplex or connectedness convention, the printed conclusion is false. Take $X=\{x_0,x_1\}$ covered by its two singleton subcomplexes and $h^{\ast }=H^{\ast }(-;\mathbb{Z})$. Both subcomplexes are contractible, but $H^0(X,x_0)$ contains the function $w$ with values $0,1$. Pointwise multiplication gives $w^2=w\ne 0$. This states precisely the needed interpretation rather than assuming a contractible component lies in the basepoint component.

Use integral homology throughout. The singular chain group $C_k(X)$ is free on the continuous maps from the standard $k$-simplex to $X$. Since the boundary operator preserves chains lying in $A$, these form a chain subcomplex. The relative chain complex and its relative homology areThe short exact sequence of chain complexes gives the long exact sequence in relative homology.

A sufficient hypothesis for the quotient comparison is a nonempty good pair: $A$ is closed and has an open neighbourhood $U$ which admits a deformation retraction onto $A$ while fixing $A$ throughout. A CW pair is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular collar neighbourhood. The induced quotient map $q:X\to Y=X/A$ collapses $A$ to a point $a$.

Here is the proof of the collapsing a pair theorem. Put $V=U/A\subset Y$. The deformation retraction makes $H_{\ast }(U,A)=0$, so the long exact sequence of the triple gives $H_{\ast }(X,A)\cong H_{\ast }(X,U)$. It also contracts $V$ onto $a$, giving $H_{\ast }(Y,\{a\})\cong H_{\ast }(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, excision givesSimilarly, excision of the closed point $a$ in the open set $V$ givesThe quotient map restricts to a homeomorphism between the two punctured pairs, so these are the same groups and the identifications commute with $q_{\ast }$. Finally $H_{\ast }(Y,\{a\})\cong {\tilde{H}}_{\ast }(Y)$, including degree zero. ThusThe neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.

For the genus-two closed orientable surface with its chosen orientation, $H_0(X)=\mathbb{Z}$, $H_1(X)={\mathbb{Z}}^4$, $H_2(X)=\mathbb{Z}$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb{Z}$. Its relevant long exact sequence in relative homology isThe final map is an isomorphism because both spaces are connected. Consequently $H_1(X,A)=\mathrm{coker}{i}_{\ast }$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The quotient topological space is connected, so its unreduced $H_0$ is $\mathbb{Z}$.

In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its homology class in $X$ is therefore zero, so $i_{\ast }=0$. The exact sequence becomesThe first sequence splits since its quotient is a free abelian group. Hence the homology after collapsing a separating surface curve isGeometrically, collapsing the boundary of each once-punctured torus fills its puncture with a cone on the circle, which is a disk. The quotient is a wedge sum of two tori at the collapsed point. Their two independent fundamental classes explain the extra second-homology generator.

In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed intersection pairing on an oriented surface therefore provides an integer homomorphism $H_1(X)\to \mathbb{Z}$ taking $[A]$ to $\pm 1$. Thus $[A]$ is a nonzero primitive homology class, and $i_{\ast }:\mathbb{Z}\to {\mathbb{Z}}^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is ${\mathbb{Z}}^3$. The same exact sequence gives $H_2(X,A)\cong H_2(X)\cong \mathbb{Z}$. Therefore the homology after collapsing a nonseparating surface curve isAnother description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed torus with two marked points, then identify the two points. Identifying two distinct points in a connected CW complex adds a loop up to homotopy, so this quotient has the homotopy type of $T^2\vee S^1$. The preceding relative homology calculation proves the groups without requiring that homotopy description.

A Serre fibration $p:E\to B$ has the homotopy lifting property for every disc: given $h:D^k\times I\to B$ and ${\tilde{h}}_0:D^k\to E$ with $p{\tilde{h}}_0=h(-,0)$, there is a lift $\tilde{h}$ extending ${\tilde{h}}_0$. Equivalently one may test this property on CW complexes.

For a map $g:E\to B$, its mapping-path fibration replacement isPath spaces have the compact-open topology, with the usual compactly generated convention. The inclusion $s(e)=(e,{\mathrm{const}}_{g(e)})$ satisfies $ps=g$. Forgetting the path is a homotopy inverse to $s$: shorten $\gamma$ to $u\mapsto \gamma (tu)$, from its constant initial path at $t=0$ to the original path at $t=1$. The equivalence here means a commuting triangle over the same base whose map of total spaces is a homotopy equivalence; it does not assert a fibrewise homotopy inverse over the base. Appending the successive segments of a moving base path gives the homotopy lifting property for $p$, even for arbitrary parameter spaces, so $p$ is in particular a Serre fibration. Its ordinary fibre over $b_0$ is the homotopy fibre

For a based Serre fibration, with fibre inclusion $j:F\hookrightarrow E$, the long exact sequence of homotopy groups of a fibration isThe maps $j_{\ast },p_{\ast }$ compose based sphere representatives with $j,p$; the maps on path components do the same with points. To define $\partial$, represent a class by $(I^k,\partial I^k)\to (B,b_0)$ and lift the cube starting at $e_0$ on its bottom face, keeping its vertical sides at $e_0$. The opposite face then gives $(I^{k-1},\partial I^{k-1})\to (F,e_0)$. Relative lifting for a CW pair gives this construction; its based homotopy class is independent of choices. At $k=1$ it records the component of the endpoint of the lifted loop. The last terms are pointed sets rather than groups in general.

Now write $W=S^1\vee S^n$ and let $f:W\to S^n$ be the collapse map. Inclusion of the $n$-sphere is a section, so $f_{\ast }:{\pi }_k(W)\to {\pi }_k(S^n)$ is surjective in every positive degree. Thus all connecting maps in its homotopy fibre sequence vanish. The fibre is path-connected, andThe section also gives the first assertion when $n=2$, by killing the connecting map from ${\pi }_2(S^2)$.

The universal cover of $W$ consists of a line with an $n$-sphere attached at each integer. It is homotopy equivalent to the wedge sum ${\bigvee }_{k\in \mathbb{Z}}{S}^n$. This cover is $(n-1)$-connected, so the Hurewicz theorem identifies its $n$th homotopy group with its $n$th homology. Choose a generator $t$ of the fundamental group whose action shifts the sphere indexed by $k$ to that indexed by $k+1$. The resulting homotopy groups as modules over the fundamental group giveEvery lifted sphere maps to $S^n$ with degree one. Consequently $f_{\ast }$ is the augmentation $\epsilon (\sum a_k{t}^k)=\sum a_k$, with the trivial $R$-action on $\mathbb{Z}$. The inclusion of the fibre respects the fundamental group actions, soThe augmentation ideal of the infinite cyclic group is generated by $t-1$: multiply a Laurent polynomial by a power of $t$ and use divisibility by $t-1$ when its value at one vanishes. Multiplication by $t-1$ is injective, giving the last isomorphism. This is a free module of rank one, whose underlying abelian group has countably infinite rank; its ${\pi }_1$-action is not trivial.

For CW pairs $(X,A)$ and $(Y,B)$, write their product pair asUse the compactly generated product, equivalently the weak CW complex topology on products of cells. For arbitrary infinite CW complexes this can be finer than the ordinary product topology. The ordinary product topology already agrees when one factor is locally compact, or when both factors have countably many cells; in particular there is no distinction for finite CW complexes.^[ref]

The quotient of this pair is $(X/A)\wedge (Y/B)$. Setwise, the collapsed subset is exactly the set of pairs with at least one coordinate in its collapsed subspace; the compatible quotient constructions give the homeomorphism in the compactly generated space convention. In a multiplicative generalized cohomology theory, external products of relative classes of degrees $k,m$ lie in degree $k+m$ for this pair. Pullback along a diagonal map gives the relative cup product.