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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.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ 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 $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. 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 $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, 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 $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ 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.

If $fg\simeq 1_Y$ and $hf\simeq 1_X$, composing homotopies givesThus $gf\simeq hf\simeq 1_X$, while $fg\simeq 1_Y$ already holds, so $g$ is a homotopy inverse of $f$.

More generally let $a:Y\to Y$ be a homotopy inverse of $fg$ and $b:X\to X$ one of $hf$. Then $g^{\prime }=ga$ satisfies $f{g}^{\prime }\simeq 1_Y$, and $h^{\prime }=bh$ satisfies $h^{\prime }f\simeq 1_X$. The first argument applies. Hence the separate one-sided hypotheses imply that $f$ is a homotopy equivalence, as expressed by separate homotopy inverses combine into a homotopy equivalence.

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.

Define a simplicial map $j:K\to L$ on vertices byThe image of every simplex is a simplex of $L$, and $j\circ i={\mathrm{id}}_L$. The maps $i\circ j$ and ${\mathrm{id}}_K$ are contiguous simplicial maps, since the union of their images on every simplex is contained in a simplex of $K$. Their realizations are therefore homotopic, so $|j|$ is a homotopy inverse of $|i|$.

Here is an explicit chain homotopy. Put $f=i_{\bullet }{j}_{\bullet }$ and use the orientations from part (b). DefineWe verifyIt is immediate on every vertex except $v_3$, where $\partial h_0(v_3)=v_3-v_1=v_3-f(v_3)$. It is immediate on the fixed edges. For the remaining edges,because $f(e_{23})=[v_2{v}_1]=-e_{12}$. Finally,since $f(t)$ is degenerate and hence zero. Thus $h$ is the required chain homotopy.

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

For points $p,q$ in a topological space $X$, the fixed-endpoint path space iswith the subspace compact-open topology from the continuous-map space. For a metric space, this agrees with uniform convergence of paths on $[0,1]$. If $p,q$ lie in different path components, this path space is empty.

If $p,q$ lie in the same path component, choose one continuous path $\alpha$ from $p$ to $q$. Appending the reversed $\alpha$ sends a path from $p$ to $q$ to a based loop at $p$. Conversely, append $\alpha$ to a based loop. The backtracking pieces contract through paths with fixed endpoints, so these maps are homotopy inverses. Hence $\Omega (p,q)$ has the homotopy type of the based loop space ${\Omega }_{p}X$. In the usual smooth manifold setting, smooth, piecewise smooth and Sobolev $H^1$ path models give the same homotopy type.