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Glue the boundary of a Möbius band to the projective line in the real projective plane by a mapping degree $n$. The band is the mapping cylinder of the degree-two boundary covering of its core. Collapsing the connecting tree in the resulting double mapping cylinder gives a CW complex with one vertex, edges $a,b$, and two attaching words $a^2$ and $a^{-n}{b}^2$. Its cellular chain complex has degree-two boundary matrix $\left(\begin{array}{cc}2& -n\\ 0& 2\end{array}\right)$. Its determinant is four, so $H_2=0$. The Smith normal form gives $H_1\cong \mathbb{Z}/4$ for odd $n$ and $(\mathbb{Z}/2{)}^2$ for even $n$; $H_0=\mathbb{Z}$ and higher homology vanishes. The space cannot be homotopy equivalent to a closed manifold: mod-two homology forces a potential manifold to be a surface, and its Euler characteristic is one, whereas the only such closed connected surface is the projective plane, with different integral $H_1$.

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.

Use the mapping cylinder convention $(x,1)\sim f(x)$ for $f:X\to Y$. Its bottom copy of $X$ is a retract precisely when $f$ has a left homotopy inverse $g:Y\to X$. A retraction restricted to $Y$ supplies $g$, and its values along the cylinder supply the homotopy from ${\mathrm{id}}_X$ to $gf$. Conversely use that homotopy on the cylinder and $g$ on $Y$; agreement at the attached end makes the map descend by the universal property of the quotient topology.

For a continuous self-map $f:X\to X$, the mapping torus has the Wang sequence even when $f$ is not a homeomorphism. Cover the time circle by a seam neighborhood and an interval away from the seam. Their inverse images retract to $X$: the seam neighborhood is a mapping cylinder, and the other is a product. Their intersection retracts to two copies of $X$. The Mayer–Vietoris sequence map on homology is, after consistent choices of signs,Invertible changes of coordinates reduce it to the direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the isomorphism summands gives the Wang sequence. This proof does not assume a fiber bundle structure for a noninvertible $f$.

Use the unit $\eta$ just constructed and replace it by the inclusion $A=S^{r-1}\hookrightarrow Y=\mathrm{Cyl}(\eta )$ into its mapping cylinder; $Y\simeq \Omega S^r$. Put $m=2r-3$. The homology isomorphisms from question 4(ii) and the relative exact sequence give $H_j(Y,A;\mathbb{Z})=0$ for $j\le m$.

Both $A$ and $Y$ are simply connected for $r\ge 3$. For the degree-two start, the absolute Hurewicz theorem identifies their ${\pi }_2$ with $H_2$; naturality and the homology isomorphism show ${\eta }_{\ast }$ is an isomorphism on ${\pi }_2$, so ${\pi }_2(Y,A)=0$. The relative fundamental group also vanishes. If there were a first nonzero relative homotopy group in some degree $3\le k\le m$, the Relative Hurewicz theorem for a $(k-1)$-connected simply connected pair would identify it with $H_k(Y,A)$, which is zero. Induction therefore gives ${\pi }_j(Y,A)=0$ for all $j\le m$.

The relative homotopy sequence now makes ${\eta }_{\ast }:{\pi }_i(S^{r-1})\to {\pi }_i(\Omega S^r)$ an isomorphism when $i+1\le m$, and a surjection at $i=m$. Under loop-suspension adjunction, the adjoint of $\eta \circ a$ is $\Sigma a$, so this particular isomorphism is the suspension homomorphism. Consequently the requested Freudenthal suspension theorem isFor $i=0$ this is read as the identification of connected-component sets; all degree-one groups here also vanish. The proof is the homology comparison proof of Freudenthal suspension.

To compute the specified unstable groups, $S^1$ has contractible universal cover, so ${\pi }_2(S^1)=0$. The Hopf fibration $S^1\to S^3\to S^2$ has exact segment $0={\pi }_3{S}^1\to {\pi }_3{S}^3\to {\pi }_3{S}^2\to {\pi }_2{S}^1=0$, giving ${\pi }_3{S}^2=\mathbb{Z}$. The supplied starting value is ${\pi }_4{S}^3=\mathbb{Z}/2$. To pass from the group over $S^{n-1}$ to that over $S^n$, use $r=n,i=n$; the condition $n\le 2n-4$ holds exactly when $n\ge 4$. Thus every later step is an isomorphism, andThis is the first homotopy group above the dimension of a sphere. The stable range deliberately excludes the step from $n=2$ to $n=3$, so the different groups there do not conflict with the theorem.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

For an open cover $X=A\cup B$, the Mayer–Vietoris theorem gives the long exact sequenceThe same sequence holds for a union of simplicial subcomplexes. All homology groups here have integer coefficients.

An explicit triangulation of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\dots ,a_{3q-1}$, and a cone vertex $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include trianglesfor every $j$, together with all their faces. The first two types triangulate the mapping cylinder of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine simplicial complex, not just a one-cell presentation.

Let $A$ be the mapping cylinder and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces toConnectedness and the absence of higher-dimensional simplices giveThis is a two-dimensional Moore space with torsion in degree one.

The suspension of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris givesIn particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. ThenDecompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb{Z}/q$ by $S^{i-1}{X}_q$, meaning $i-1$ successive suspensions. Take their finite wedge, or a point if there are no summands. The suspension formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected polyhedron satisfies $\boxed{H_0(X)=\mathbb{Z},H_i(X)=G_i(1\le i\le n),H_i(X)=0(i>n)}$. The finite triangulations above, their simplicial suspensions and wedges provide an actual polyhedron.

Let ${\mathcal{C}}_{\mathrm{fg}}$ be the Serre class of finitely generated abelian groups. A homomorphism is an isomorphism modulo this class when its kernel and cokernel are finitely generated. The Hurewicz theorem modulo a Serre class says that, for a simply connected space and $r\ge 2$, if ${\pi }_i(X)\in {\mathcal{C}}_{\mathrm{fg}}$ for $2\le i<r$, thenThe corresponding lower-dimensional homology condition is equivalent to the lower homotopy condition. In particular, degreewise finite generation of integral homology and homotopy are equivalent for simply connected spaces. The Serre class is closed under subgroups, quotients and extensions; these closure properties are what make the modulo-class formulation useful.

Here is the consequence needed for construction. Inductively, if the lower homotopy groups are finitely generated, the theorem gives a finitely generated kernel for $h_r$. Its image is a subgroup of the finitely generated $H_r(X)$, so the image is also finitely generated. The resulting extension proves ${\pi }_r(X)$ finitely generated. Starting with the ordinary isomorphism ${\pi }_2(X)\cong H_2(X)$ yields this for every positive degree.

We now construct a finite type CW approximation. Choose finitely many maps $S^2\to X$ generating ${\pi }_2(X)$, and let $C_2$ be their wedge. The resulting map $C_2\to X$ is $2$-connected: it is an isomorphism below degree two and a surjection in degree two. In general, call a map $r$-connected when its mapping-cylinder pair has relative homotopy groups zero through degree $r$.

Suppose $f_r:C_r\to X$ is $r$-connected and $C_r$ is a finite simply connected CW complex of dimension at most $r$. The homotopy groups of $C_r$ are finitely generated by the same modulo-class theorem, since a finite CW complex has finitely generated homology. The long exact sequence of relative homotopy groups shows thatis finitely generated: it lies between a quotient of ${\pi }_{r+1}(X)$ and a subgroup of ${\pi }_r(C_r)$. Relative groups here refer to the mapping cylinder of $f_r$.

Represent a finite set of generators by relative disks. Attach their boundary spheres to $C_r$, and extend the map over the disks by their chosen maps into $X$. This adds finitely many $(r+1)$-cells and kills the relative group in that dimension without changing lower relative groups. The new map $C_{r+1}\to X$ is $(r+1)$-connected. Iterating givesThere are finitely many cells of each dimension. Every fixed homotopy degree stabilizes to an isomorphism once sufficiently high-dimensional cells have been added, so this is the required weak equivalence:

For the bounded-homology assertion, assume first $n\ge 2$ and take the finite $n$-dimensional $C_n$ just constructed. The Relative Hurewicz theorem for its $n$-connected map givesThe relative homology sequence and $H_{n+1}(X)=0$ identify the latter withBecause $C_n$ is $n$-dimensional, $H_n(C_n)$ is a subgroup of its free cellular $n$-chain group. It is finite free, and so is $K$.

Choose a basis of $K$, lift it using the Relative Hurewicz theorem, and attach exactly those finitely many $(n+1)$-cells to $C_n$. Denote the resulting complex by $C^{\prime }$. Its new cellular boundary has image $K$ and is injective: the selected cycles are linearly independent in $H_n(C_n)$, and there are no old $(n+1)$-boundaries. ThusLower homology remains unchanged, and all higher homology is zero on both sides. The map $C^{\prime }\to X$ is an integral homology isomorphism between simply connected spaces. The homological Whitehead theorem consequently makes it a weak homotopy equivalence. Hence the finite CW approximation from bounded homology hasIf $n=0$ or $1$, simple connectivity and the homology hypothesis make all reduced homology zero. The Hurewicz theorem, applied at the first possible nonzero homotopy degree, shows that $X$ is weakly contractible, so a point suffices. If the wording requires dimension exactly $n+1$ rather than at most $n+1$, add a contractible cancelling pair of $n$- and $(n+1)$-cells, mapping constantly to the basepoint. This does not change the weak homotopy type.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

Let $i:Y\hookrightarrow M_f$ be the canonical inclusion and defineThis is well-defined because $[0,x]=f(x)$ in the mapping cylinder, and continuity follows from the universal property of the quotient topology. Clearly $r\circ i={\mathrm{id}}_Y$.

DefineThe formula respects the gluing relation at $t=0$, so it descends from the disjoint union to a continuous homotopy. It satisfies $H_0={\mathrm{id}}_{M_f}$ and $H_1=i\circ r$. Thusindeed $Y$ is a deformation retract of $M_f$.

Cut the square model of the Klein bottle along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a Möbius band, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The Seifert-van Kampen theorem therefore gives

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentationThe orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroupBy the classification of connected covering spaces, $H$ determines the orientation double cover of the Klein bottleChoosing torus generators $\alpha ,\beta$ along the two translation directions, we may takeThese commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the mapping cylinder of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/\sim \cong K$. Hence

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, toThe intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to ${\pi }_1(U)\cong {\pi }_1(K)$ has image preciselyThe orientation homomorphismis zero on this intersection, so it is compatible with the trivial homomorphism from ${\pi }_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjectionThereforeThis is the Klein-bottle mapping-cylinder obstruction to simple connectivity.

Solutions to all three questions of the topology and groups course of the University of Oxford Part B examination, sat on 31 May 2023. Source: original released PDF, bundle pages 221–223; paper pages 1–3. The Mathematics and Statistics version is labelled B3.4. All eighteen parts are treated below. Question 3(a)(iii), as printed, needs an additional hypothesis such as local path connectedness; its solution includes both the intended argument and a counterexample to the unrestricted assertion.

Throughout Question 1, $I=[0,1]$ has its usual topology, and the mapping cylinder uses the paper's convention that $(x,1)$ is attached to $f(x)$.

DefineThe two formulas agree on the attached end, so the universal property of the quotient topology makes $r$ a continuous map. Also $ri={\mathrm{id}}_Y$.

Slide each cylinder point towards its attached end usingAt $t=1$ these formulas both give $i(f(x))$, so they define a homotopy on the mapping cylinder. For continuity, the product of a quotient map with the unit interval is a quotient map; the formulas before taking the quotient topology are continuous maps and agree on every identified pair. ThusMoreover $Y$ stays fixed throughout. We have obtained the stronger conclusionIn particular $r$ is the requested homotopy retract.

Writefor the quotient map defining the mapping cylinder, and put $j(x)=q(x,0)$. Since the equivalence relation only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an injective continuous map.

Define the homotopyIt is a continuous map, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. HenceEven if $f$ identifies many points, the free end of the mapping cylinder retains a separate copy of every point of $X$.

Based relative maps of a disk into $X$ with its boundary mapped into $A$, modulo homotopies preserving these conditions. Replacing a map by its mapping cylinder gives relative groups for a general map. They fit into the long exact sequence of relative homotopy groups and detect connectivity of the map.