Past exam of the topology and groups course of the University of Oxford / 2023 / 1

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.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Let $i:A\hookrightarrow X$ be the inclusion map and $r:X\to A$ the retraction. Since $X$ is contractible, choose a contraction of a topological spaceThe function compositionis a continuous map. Its endpoints satisfyConsequently $K$ is a homotopy from the identity map of $A$ to a constant function. This provesThe contracting point of $X$ need not belong to $A$: applying the retraction sends it to the contracting point $r(x_0)$ of $A$. This is the retract of a contractible space principle.

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 $X$.

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.

Identify the bottom $j(X)=q(X\times \{0\})$ with $X$; its natural map $j$ is a topological embedding. To verify this directly, an open subset $O\subseteq X$ gives the saturated open subset $O\times [0,1/2)$ of the space before taking the quotient topology; its image is open in $M(f)$ and meets the bottom in $j(O)$.

Suppose first that there is a retraction $R:M(f)\to j(X)$. Define the continuous mapThenis a homotopy with $L(x,0)=x$ and $L(x,1)=g(f(x))$. Hence $gf\simeq {\mathrm{id}}_X$.

Conversely, suppose a continuous map $g:Y\to X$ has $gf\simeq {\mathrm{id}}_X$. Reverse the parameter if necessary to choose a homotopy $L$ with $L(x,0)=x$ and $L(x,1)=gf(x)$. SetAt the glued end, $j(L(x,1))=j(gf(x))$, exactly the value prescribed at $i(f(x))$. The universal property of the quotient topology therefore gives a continuous map $R:M(f)\to j(X)$. At the bottom, $R(j(x))=j(x)$, proving that $R$ is a retraction. We have established the mapping-cylinder bottom retraction criterion: