# General Topology/Homotopy

Definition (homotopy):

Let ${\displaystyle X,Y}$ be topological spaces, and let ${\displaystyle f,g:X\to Y}$ be two continuous maps. A homotopy between ${\displaystyle f}$ and ${\displaystyle g}$ is a continuous map

${\displaystyle H:[0,1]\times X\to Y}$, such that: ${\displaystyle \forall x\in X:H_{0}(x)=f(x),H_{1}(x)=g(x)}$.

(Note that the first argument of ${\displaystyle H}$ is usually denoted in subscript.) In the case of the existence of such an ${\displaystyle H}$, the functions ${\displaystyle f}$ and ${\displaystyle g}$ are said to be homotopic to each other.

This means that ${\displaystyle H}$ is a "continuous transformation" of ${\displaystyle f}$ into ${\displaystyle g}$ or vice versa.

Definition (relative homotopy):

Let ${\displaystyle X}$ be a topological space, and let ${\displaystyle A\subseteq X}$ be a subset. If ${\displaystyle Y}$ is another topological space, and if ${\displaystyle f,g:X\to Y}$ are continuous functions, then ${\displaystyle f}$ and ${\displaystyle g}$ are said to be homotopic relative ${\displaystyle A}$ (in short: ${\displaystyle f\simeq g\operatorname {rel} A}$) if there exists a homotopy ${\displaystyle H:I\times X\to Y}$ between ${\displaystyle f}$ and ${\displaystyle g}$ such that ${\displaystyle H_{t}(a)=f(a)=g(a)}$ for all ${\displaystyle (t,a)\in I\times A}$.

Proposition (relative homotopy is an equivalence relation):

If ${\displaystyle X}$ is a topological space, ${\displaystyle A\subseteq X}$ a subset and ${\displaystyle Y}$ another topological space, then the relation

${\displaystyle f\simeq g\operatorname {rel} A}$

is an equivalence relation on ${\displaystyle {\mathcal {C}}(X,Y)}$.

Proof: First, note that by defining ${\displaystyle H_{t}(x):=f(x)}$ for ${\displaystyle (t,x)\in [0,1]\times X}$, that ${\displaystyle H}$ is continuous, since ${\displaystyle H^{-1}(U)=[0,1]\times f^{-1}(U)}$ for ${\displaystyle U\subseteq Y}$ open, from which we gain reflexivity as also ${\displaystyle f(a)=H_{t}(a)}$ whenever ${\displaystyle a\in A}$. Symmetry follows upon considering, given a homotopy ${\displaystyle H}$ from ${\displaystyle f}$ to ${\displaystyle g}$, the homotopy ${\displaystyle {\overline {H}}_{t}(x):=H_{1-t}(x)}$, which is continuous as the composition of ${\displaystyle H}$ with the continuous function ${\displaystyle (t,x)\mapsto (1-t,x)}$; indeed, the latter map is continuous since both of its components are; it is also a homotopy relative to ${\displaystyle A}$, since ${\displaystyle H_{1-t}(a)=f(a)=g(a)}$ for ${\displaystyle a\in A}$. Finally, note that transitivity holds, since if ${\displaystyle f\sim g}$ via the homotopy ${\displaystyle H}$ and ${\displaystyle g\sim h}$ via the homotopy ${\displaystyle H'}$, then ${\displaystyle f\sim h}$ via the homotopy

${\displaystyle H*H'(t,x):={\begin{cases}H(2t,x)&t\leq {\frac {1}{2}}\\H'(2t-1,x)&t\geq {\frac {1}{2}}\end{cases}}}$,

which is continuous since it is continuous on two closed sets that make up the whole space and relative to ${\displaystyle A}$ since both ${\displaystyle H}$ and ${\displaystyle H'}$ are relative to ${\displaystyle A}$. ${\displaystyle \Box }$

Definition (retract):

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle A}$ be a subset of ${\displaystyle X}$. A retract of ${\displaystyle X}$ onto ${\displaystyle A}$ is a continuous function ${\displaystyle r:X\to A}$ so that ${\displaystyle r|_{A}=\operatorname {Id} _{A}}$, the identity.

Definition (deformation retraction):

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle A}$ be a subset of ${\displaystyle X}$. A deformation retraction of ${\displaystyle X}$ onto ${\displaystyle A}$ is a continuous function ${\displaystyle r:[0,1]\times X\to X}$ so that ${\displaystyle r_{0}(x)=x}$ for all ${\displaystyle x\in X}$ (that is, ${\displaystyle r_{0}=\operatorname {Id} _{X}}$) and ${\displaystyle r_{1}}$ is a retract onto ${\displaystyle A}$.

Note that the first argument is denoted in the lower index, so that instead of ${\displaystyle r(t,x)}$ we write ${\displaystyle r_{t}(x)}$.

That is, a deformation retraction is a homotopy between the identity and a retract onto ${\displaystyle A}$.

Definition (homotopy equivalence):

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be topological spaces. A homotopy equivalence between ${\displaystyle X}$ and ${\displaystyle Y}$ is a pair of continuous functions ${\displaystyle f:X\to Y}$ and ${\displaystyle g:Y\to X}$ so that ${\displaystyle g\circ f\simeq \operatorname {Id} _{X}}$ and ${\displaystyle f\circ g\simeq \operatorname {Id} _{Y}}$.

Proposition (deformation retraction induces homotopy equivalence):

Let ${\displaystyle X}$ be a topological space, let ${\displaystyle A\subseteq X}$ be a subset, and let ${\displaystyle r:[0,1]\times X\to X}$ be a deformation retraction of ${\displaystyle X}$ onto ${\displaystyle A}$. Then a homotopy equivalence ${\displaystyle X\simeq A}$ is given by the pair ${\displaystyle r_{1}:X\to A}$ and ${\displaystyle \iota :A\to X}$, ${\displaystyle \iota (a)=a}$ being the inclusion.

Proof: We have ${\displaystyle r_{1}\circ \iota =\operatorname {Id} _{A}}$, which is equal (and hence homotopic) to ${\displaystyle \operatorname {Id} _{A}}$. Further, a homotopy between ${\displaystyle \operatorname {Id} _{X}}$ and ${\displaystyle \iota \circ r_{1}=r_{1}}$ is given by ${\displaystyle r}$ itself. ${\displaystyle \Box }$

Definition (contractible):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called contractible iff there exists a point ${\displaystyle x_{0}\in X}$ so that ${\displaystyle X}$ deformation retracts onto ${\displaystyle \{X_{0}\}}$.

Proposition (contractible spaces are path-connected):

Let ${\displaystyle X}$ be a contractible space, that contracts to a point ${\displaystyle x_{0}\in X}$. Then any two points ${\displaystyle x,y\in X}$ can be connected by a path that runs through ${\displaystyle x_{0}}$.

Proof: For any point ${\displaystyle x\in X}$, if ${\displaystyle r:[0,1]\times X\to X}$ is the deformation retraction of ${\displaystyle X}$ onto ${\displaystyle \{x_{0}\}}$, we get a path from ${\displaystyle x}$ to ${\displaystyle x_{0}}$ by the way of

${\displaystyle \gamma :[0,1]\to X,\gamma (t):=r_{t}(x)}$,

which is continuous as the composition of continuous functions, since the map ${\displaystyle t\mapsto (t,x)}$ is continuous, since the first component is the identity and the second the constant function. ${\displaystyle \Box }$

Definition (nullhomotopic):

Let ${\displaystyle f:X\to Y}$ be a continuous function between topological spaces. ${\displaystyle f}$ is said to be nullhomotopic iff it is homotopic to a constant function, that is, to a function ${\displaystyle g:X\to Y}$ such that there exists ${\displaystyle y_{0}\in Y}$ so that ${\displaystyle g(x)=y_{0}}$ for all ${\displaystyle x\in X}$.

Proposition (characterisation of contractibility):

Let ${\displaystyle X}$ be a topological space. The following are equivalent:

1. ${\displaystyle X}$ is contractible
2. The identity map on ${\displaystyle X}$ is nullhomotopic
3. [/itex]X[/itex] is homotopy equivalent to a one-point space
4. For every topological space ${\displaystyle Z}$, every continuous function ${\displaystyle f:Z\to X}$ is nullhomotopic
5. For every topological space ${\displaystyle Z}$, any two continuous functions ${\displaystyle f:Z\to X}$ and ${\displaystyle g:Z\to X}$ are homotopic

Proof: Suppose first that ${\displaystyle X}$ is contractible, and let ${\displaystyle r:[0,1]\times X\to X}$ be the contraction. Then ${\displaystyle r}$ is simultaneously a homotopy between the identity and a constant map, where the latter has the value of the point onto which ${\displaystyle X}$ contracts. If the identity is homotopic to a constant map, and ${\displaystyle H:[0,1]\times X\to X}$ is the respective homotopy, then ${\displaystyle H}$ is also a deformation retraction and we conclude 3. since this deformation retraction induces a homotopy equivalence between a one-point space and ${\displaystyle X}$. Let now ${\displaystyle Z}$ be a topological space and ${\displaystyle f:Z\to X}$ a continuous function. Suppose further that ${\displaystyle g:X\to \{x_{0}\}}$ and ${\displaystyle h:\{x_{0}\}\to X}$ constitute a homotopy equivalence. Then in particular ${\displaystyle h\circ g\simeq \operatorname {Id} _{X}}$ via some homotopy ${\displaystyle H}$. But ${\displaystyle h\circ g}$ is a constant map. We then get ${\displaystyle f\simeq h\circ g\circ f}$ via the homotopy ${\displaystyle {\tilde {H}}(t,z):=H(t,f(z)}$, so that ${\displaystyle f}$ is nullhomotopic. Suppose now that ${\displaystyle f:Z\to X}$ and ${\displaystyle g:Z\to X}$ are two continuous functions. They will both be nullhomotopic, and if we can show that ${\displaystyle X}$ is path-connected and ${\displaystyle {\tilde {f}}}$ is the constant function to which ${\displaystyle f}$ is homotopic and ${\displaystyle x_{0}:={\tilde {f}}(z)}$ for any ${\displaystyle z}$, then they will both be homotopic to ${\displaystyle {\tilde {f}}}$, since ${\displaystyle {\tilde {f}}}$ is homotopic to any constant map via ${\displaystyle H(t,x):=(z\mapsto \gamma (t))}$, where ${\displaystyle \gamma }$ is a curve from ${\displaystyle x_{0}}$ to the point of the other constant map ${\displaystyle Z\to X}$. But ${\displaystyle X}$ is path-connected, since upon choosing ${\displaystyle Z:=\{z_{0},z_{1}\}}$, the two-point space with discrete topology, we get that the (continuous) function sending ${\displaystyle z_{0}}$ and ${\displaystyle z_{1}}$ to two points ${\displaystyle x_{0},x_{1}\in X}$ is nullhomotopic, and the homotopy ${\displaystyle H}$ yields a path between ${\displaystyle x_{0}}$ and ${\displaystyle x_{1}}$ by the way of

${\displaystyle \gamma :=\rho _{0}*{\overline {\rho _{1}}}}$, where ${\displaystyle \rho _{0}(t):=H(t,z_{0})}$ and ${\displaystyle \rho _{1}(t):=H(t,z_{1})}$.

Finally, if any two continuous functions from any space to ${\displaystyle X}$ are homotopic, we may as well choose ${\displaystyle Z:=X}$ and one map to be any constant map and the other one the identity; the homotopy between the two then yields the desired deformation retraction. ${\displaystyle \Box }$

Definition (homotopy extension property):

Let ${\displaystyle X}$ be a topological space and ${\displaystyle A\subseteq X}$ a subset. The pair ${\displaystyle (X,A)}$ is said to have the homotopy extension property if and only if for every topological space ${\displaystyle Y}$ and every continuous function ${\displaystyle f:X\to Y}$ such that ${\displaystyle f|_{A}}$ is homotopic to some other map ${\displaystyle H_{1}}$ via a homotopy ${\displaystyle H}$, there exists a homotopy ${\displaystyle {\tilde {H}}}$ so that ${\displaystyle {\tilde {H}}_{0}=f}$ and ${\displaystyle {\tilde {H}}|_{[0,1]\times A}=H}$.

Proposition (characterisation of the homotopy extension property via retraction):

Let ${\displaystyle X}$ be a topological space and ${\displaystyle A\subseteq X}$ a subset. ${\displaystyle (X,A)}$ has the homotopy extension property if and only if there exists a retraction from ${\displaystyle [0,1]\times X}$ onto ${\displaystyle \{0\}\times X\cup [0,1]\times A}$.

Proof: Suppose first that ${\displaystyle (X,A)}$ does have the homotopy extension property. Then pick ${\displaystyle Y:=\{0\}\times X\cup [0,1]\times A}$ and define ${\displaystyle f:X\to Y}$ by ${\displaystyle f(x)=(0,x)}$. Then ${\displaystyle f|_{A}}$ is homotopic to the function ${\displaystyle g(a)=(1,a)}$ via ${\displaystyle H(t,x)=(t,x)}$, and then by the homotopy extension property we get a homotopy ${\displaystyle {\tilde {H}}}$ of ${\displaystyle f}$ to some other function ${\displaystyle {\tilde {H}}}$. By the form of ${\displaystyle H}$, ${\displaystyle {\tilde {H}}}$ restricts to the identity on ${\displaystyle \{0\}\times X\cup [0,1]\times A\subseteq [0,1]\times X}$, and further it maps ${\displaystyle [0,1]\times X}$ to ${\displaystyle \{0\}\times X\cup [0,1]\times A}$, so that ${\displaystyle {\tilde {H}}}$ is a retraction. Suppose now that ${\displaystyle \{0\}\times X\cup [0,1]\times A}$ is a retract of ${\displaystyle [0,1]\times X}$ via the function ${\displaystyle r:[0,1]\times X\to Y}$, where we define ${\displaystyle Y:=\{0\}\times X\cup [0,1]\times A}$ to ease notation. Let ${\displaystyle f:X\to Z}$ (where ${\displaystyle Z}$ is some topological space) be a function and ${\displaystyle H:[0,1]\times A\to A}$ a homotopy so that ${\displaystyle H_{0}=f|_{A}}$. We define

${\displaystyle {\tilde {H}}:[0,1]\times X\to Z,{\tilde {H}}(t,x):={\begin{cases}f(x)&r(t,x)\in \{0\}\times X\\{\tilde {H}}(r(t,x))&r(t,x)\in [0,1]\times A\end{cases}}}$

and claim that it is continuous. Note that ${\displaystyle {\tilde {H}}}$ is the composition of ${\displaystyle r}$ (which is continuous) with the map

${\displaystyle \eta :Y\to Z,\eta (t,x):={\begin{cases}f(x)&t=0\\H(t,x)&t>0\end{cases}}}$,

so that it suffices to prove that ${\displaystyle \eta }$ is continuous. To this end, it will be sufficient to prove that whenever ${\displaystyle U\subseteq Y}$ is a set such that ${\displaystyle U\cap \{0\}\times X}$ and ${\displaystyle U\cap [0,1]\times A}$ are open, then ${\displaystyle U}$ itself is open; indeed, if this is the case, then whenever ${\displaystyle V}$ is an open subset of ${\displaystyle Z}$, we'll have that

${\displaystyle \eta ^{-1}(V)=\eta ^{-1}(V)\cap \{0\}\times X\cup \eta ^{-1}\cap [0,1]\times A=\eta |_{\{0\}\times X}^{-1}(V)\cup \eta |_{[0,1]\times A}^{-1}(V)}$

is open. So suppose that ${\displaystyle U\subseteq Y}$ has said property. We show that ${\displaystyle U}$ is open by fixing a ${\displaystyle (t,x)\in U}$ and finding an open neighbourhood of ${\displaystyle (t,x)}$ in ${\displaystyle Y}$ that is contained within ${\displaystyle U}$. If ${\displaystyle (t,x)\in (0,1]\times A}$, then we may just choose a suitable product neighbourhood since cylinders are a basis of the product topology and by the definition of the subspace topology. If ${\displaystyle t=0}$ and ${\displaystyle x\notin {\overline {A}}}$, we may just choose the set ${\displaystyle U\cap \{0\}\times X}$. The only remaining case is the case ${\displaystyle x\in {\overline {A}}}$ and ${\displaystyle t=0}$. If that is the case, suppose first that ${\displaystyle s>0}$, and consider the point ${\displaystyle (s,x)\in [0,1]\times X}$. We write the retraction ${\displaystyle r:[0,1]\times X\to Y}$ as ${\displaystyle r(s,x)=(r_{1}(s,x),r_{2}(s,x))}$. Since ${\displaystyle x\in {\overline {A}}}$ and ${\displaystyle r_{1}(s,y)=s}$ for all ${\displaystyle y\in A}$, we get that ${\displaystyle r_{1}(s,x)=s}$, since denoting the neighbourhood filter of ${\displaystyle (s,x)}$ by ${\displaystyle N(s,x)}$, we get that ${\displaystyle r_{1}(N(s,x))}$ converges to ${\displaystyle r_{1}(s,x)}$ by continuity of ${\displaystyle r_{1}}$ (which follows from the characterisation of continuity of functions to a space with initial topology), but ${\displaystyle s}$ is contained in each set ${\displaystyle r_{1}(W)}$, ${\displaystyle W}$ being a neighbourhood of ${\displaystyle (s,x)}$, since ${\displaystyle W}$, by definition of the product topology, always contains a point of the form ${\displaystyle (s,y)}$, where ${\displaystyle y\in A}$, and we conclude since ${\displaystyle [0,1]}$ is Hausdorff.

For each ${\displaystyle n\in \mathbb {N} }$, we define ${\displaystyle O_{n}}$ to be the maximal open subset of ${\displaystyle X}$ so that ${\displaystyle [0,1/n)\times (O_{n}\cap A)\subseteq U}$; a maximal such set exists since sets having this property are closed under union, and we take the union over all the sets having this property. Then define

${\displaystyle O:=\bigcup _{n\in \mathbb {N} }O_{n}}$;

${\displaystyle O}$ is then an open subset of ${\displaystyle X}$. Now ${\displaystyle x\in O}$, since otherwise, for all ${\displaystyle n\in \mathbb {N} }$ we have ${\displaystyle x\notin O_{n}}$. However, then for arbitrary ${\displaystyle s>0}$ and ${\displaystyle n\in \mathbb {N} }$, we get that ${\displaystyle r_{2}(s,x)\notin O_{n}}$, since if ${\displaystyle r_{2}(s,x)\in O_{n}}$, then by continuity of ${\displaystyle r_{2}}$, we find ${\displaystyle \epsilon >0}$ and ${\displaystyle V\subseteq X}$ open with ${\displaystyle x\in V}$, so that ${\displaystyle r_{2}((s-\epsilon ,s+\epsilon )\times V)\subseteq O_{n}}$, which implies in particular that ${\displaystyle r_{2}(\{s\}\times (V\cap A)\subseteq O_{n}}$. However, for ${\displaystyle v\in V\cap A}$, we simply have ${\displaystyle r_{2}(s,v)=v}$ since ${\displaystyle r}$ is a retraction, and hence we get ${\displaystyle r_{2}(\{s\}\times (V\cap A)=V\cap A}$ and ${\displaystyle V\cap A\subseteq O_{n}}$. Hence, ${\displaystyle V\subseteq O_{n}}$ since ${\displaystyle (V\cap A)\times [0,1/n)\subseteq (O_{n}\cap A)\times [0,1/n)}$, and thus ${\displaystyle x\in V}$ implies ${\displaystyle x\in O_{n}}$. However, note that still ${\displaystyle r_{2}(s,x)\in A}$; indeed, by what was proven above, we have ${\displaystyle r_{1}(s,x)=s>0}$, and then, since ${\displaystyle r}$ is a retraction whence ${\displaystyle r(s,x)\in Y}$, we must have ${\displaystyle r_{2}(s,x)\in A}$. We conclude that ${\displaystyle r_{2}(s,x)\in A\setminus O}$ for all ${\displaystyle s>0}$. Now we observe that if ${\displaystyle \pi _{X}}$ is the projection from ${\displaystyle I\times X}$ to ${\displaystyle X}$, then ${\displaystyle \pi _{X}(U\cap \{0\}\times X)\cap A\subseteq O}$, since whenever ${\displaystyle (x,0)\in U}$ so that ${\displaystyle x\in A}$, we find an open ${\displaystyle W\subseteq A}$ and an ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle (x,0)\in [0,1/n)\times W\subseteq U}$, and we then write ${\displaystyle W=A\cap {\tilde {W}}}$, where ${\displaystyle {\tilde {W}}}$ is open, and obtain ${\displaystyle x\in {\tilde {W}}\subseteq O_{n}\subseteq O}$. By continuity of ${\displaystyle r_{2}(\cdot ,x)}$, we then obtain ${\displaystyle r_{2}(0,x)\notin U}$, since the preimage of the complement of ${\displaystyle O}$ will be closed in ${\displaystyle [0,1]}$. But ${\displaystyle r_{2}(0,x)=x}$, so that in fact ${\displaystyle x\notin U}$, a contradiction.

Hence, choose ${\displaystyle n\in \mathbb {N} }$ so that ${\displaystyle x\in O_{n}}$; then the neighbourhood ${\displaystyle [0,1/n)\times O_{n}\cap Y}$ will do as an open neighbourhood of ${\displaystyle (0,x)}$ contained within ${\displaystyle U}$, using the definitions of the subspace and product topologies. ${\displaystyle \Box }$