Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 16 1 Solution Created 2026-10-03 Updated 2026-10-14
A chain complex consists of abelian groups and homomorphisms with . Its groups of chain cycles and chain boundaries are and ; the identity ensures . Its homology groups areFor a nonnegative complex, take for .
The singular chain group is the free abelian group on continuous maps , called singular simplices. With the inclusion of the face opposite vertex , defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving . This defines the singular chain complex and its singular homology. A continuous map induces the chain map ; it commutes with the boundary because composition commutes with restriction to faces.
To prove homotopy invariance of homology, let have endpoints . Triangulate by the simplices with ordered verticesLet be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give , and the remaining faces are with their induced signs. ThusFor a chain cycle , is a chain boundary, so on homology. If is a homotopy equivalence, its homotopy inverse therefore induces an inverse to , proving that singular homology depends only on homotopy type.
Use the standard CW complex for an orientable genus-three surface: one vertex, six edges , and a two-cell whose attaching word is . The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose to be the nonseparating curve represented by , and choose as a small circle bounding a disc in a surface patch. Then is a nonzero primitive homology class, whereas the inclusion of induces zero on . If a homotopy equivalence carried onto , its restriction would factor through and kill , contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.
There is nevertheless such a continuous map. Map the edge homeomorphically onto and all other edges to its base point. The two-cell attaching word then has winding number zero in , hence is null-homotopic and the map extends to . Compose with a homeomorphism and the inclusion of in the surface. This gives . Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.
For an unambiguous drawing, realize as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.
Past exam of the mathematics course of the University of Cambridge 2008 ii Paper 4 21F i Solution Created 2026-09-24 Updated 2026-10-14
More generally let be a homotopy inverse of and one of . Then satisfies , and satisfies . The first argument applies. Hence the separate one-sided hypotheses imply that is a homotopy equivalence, as expressed by separate homotopy inverses combine into a homotopy equivalence.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 127 1 Solution Created 2026-10-03 Updated 2026-10-14
A Serre fibration has the homotopy lifting property for every disc: given and with , there is a lift extending . Equivalently one may test this property on CW complexes.
For a map , its mapping-path fibration replacement isPath spaces have the compact-open topology, with the usual compactly generated convention. The inclusion satisfies . Forgetting the path is a homotopy inverse to : shorten to , from its constant initial path at to the original path at . 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 , even for arbitrary parameter spaces, so is in particular a Serre fibration. Its ordinary fibre over is the homotopy fibre
For a based Serre fibration, with fibre inclusion , the long exact sequence of homotopy groups of a fibration isThe maps compose based sphere representatives with ; the maps on path components do the same with points. To define , represent a class by and lift the cube starting at on its bottom face, keeping its vertical sides at . The opposite face then gives . Relative lifting for a CW pair gives this construction; its based homotopy class is independent of choices. At it records the component of the endpoint of the lifted loop. The last terms are pointed sets rather than groups in general.
Now write and let be the collapse map. Inclusion of the -sphere is a section, so 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 , by killing the connecting map from .
The universal cover of consists of a line with an -sphere attached at each integer. It is homotopy equivalent to the wedge sum . This cover is -connected, so the Hurewicz theorem identifies its th homotopy group with its th homology. Choose a generator of the fundamental group whose action shifts the sphere indexed by to that indexed by . The resulting homotopy groups as modules over the fundamental group giveEvery lifted sphere maps to with degree one. Consequently is the augmentation , with the trivial -action on . The inclusion of the fibre respects the fundamental group actions, soThe augmentation ideal of the infinite cyclic group is generated by : multiply a Laurent polynomial by a power of and use divisibility by when its value at one vanishes. Multiplication by is injective, giving the last isomorphism. This is a free module of rank one, whose underlying abelian group has countably infinite rank; its -action is not trivial.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 3 20F c Solution Created 2026-09-24 Updated 2026-09-29
Define a simplicial map on vertices byThe image of every simplex is a simplex of , and . The maps and are contiguous simplicial maps, since the union of their images on every simplex is contained in a simplex of . Their realizations are therefore homotopic, so is a homotopy inverse of .
Here is an explicit chain homotopy. Put and use the orientations from part (b). DefineWe verifyIt is immediate on every vertex except , where . It is immediate on the fixed edges. For the remaining edges,because . Finally,since is degenerate and hence zero. Thus is the required chain homotopy.
