一般拓撲/同倫

定義（同倫）:

令 $X,Y$ 是拓撲空間，令 $f,g:X\to Y$ 是兩個連續對映。在 $f$ 和 $g$ 之間的同倫是一個連續對映

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

（注意，$H$ 的第一個引數通常用下標表示。）如果存在這樣的 $H$，則函式 $f$ 和 $g$ 被稱為同倫的。

這意味著 $H$ 是 $f$ 連續地“變換”為 $g$，反之亦然。

定義（相對同倫）:

令 $X$ 為拓撲空間，令 $A\subseteq X$ 為子集。如果 $Y$ 是另一個拓撲空間，並且如果 $f,g:X\to Y$ 是連續函式，那麼 $f$ 和 $g$ 被稱為**相對於 $A$ 同倫**（簡稱：$f\simeq g\operatorname {rel} A$）如果存在一個同倫 $H:I\times X\to Y$ 在 $f$ 和 $g$ 之間，使得對於所有 $(t,a)\in I\times A$，有 $H_{t}(a)=f(a)=g(a)$。

命題（相對同倫是等價關係）:

如果 $X$ 是一個拓撲空間，$A\subseteq X$ 是一個子集，並且 $Y$ 是另一個拓撲空間，那麼關係

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

是 ${\mathcal {C}}(X,Y)$ 上的一個等價關係。

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

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

它是連續的，因為它在構成整個空間的兩個閉集上是 連續的，並且相對於 $A$，因為 $H$ 和 $H'$ 都相對於 $A$。 $\Box$

定義（收縮）:

設 $X$ 是一個拓撲空間，設 $A$ 是 $X$ 的一個子集。 $X$ 到 $A$ 的一個 **收縮** 是一個連續函式 $r:X\to A$，使得 $r|_{A}=\operatorname {Id} _{A}$，即恆等式。

定義（形變收縮）:

設 $X$ 是一個拓撲空間，設 $A$ 是 $X$ 的一個子集。 $X$ 到 $A$ 的一個 **形變收縮** 是一個連續函式 $r:[0,1]\times X\to X$，使得對於所有 $x\in X$ 都有 $r_{0}(x)=x$（即 $r_{0}=\operatorname {Id} _{X}$）並且 $r_{1}$ 是一個到 $A$ 的收縮。

注意第一個引數用下標表示，因此我們不寫$r(t,x)$，而是寫 $r_{t}(x)$。

也就是說，形變縮回是恆等對映和到$A$的縮回對映之間的同倫。

定義（同倫等價）:

設$X$ 和 $Y$ 是拓撲空間。 $X$ 和 $Y$ 之間的**同倫等價**是一對連續函式$f:X\to Y$ 和 $g:Y\to X$，使得$g\circ f\simeq \operatorname {Id} _{X}$ 且 $f\circ g\simeq \operatorname {Id} _{Y}$。

命題（形變縮回誘導同倫等價）:

設$X$ 是拓撲空間，設 $A\subseteq X$ 是一個子集，並設 $r:[0,1]\times X\to X$ 是 $X$ 到 $A$ 的形變縮回。 那麼，由一對$r_{1}:X\to A$ 和 $\iota :A\to X$ 給出同倫等價$X\simeq A$，其中 $\iota (a)=a$ 是嵌入對映。

證明： 我們有 $r_{1}\circ \iota =\operatorname {Id} _{A}$，它等於（因此同倫於）$\operatorname {Id} _{A}$。此外，$\operatorname {Id} _{X}$ 和 $\iota \circ r_{1}=r_{1}$ 之間的同倫由 $r$ 本身給出。 $\Box$

定義（可縮空間）:

令 $X$ 為拓撲空間。 $X$ 被稱為可縮空間當且僅當存在一個點 $x_{0}\in X$ 使得 $X$ 到 $\{X_{0}\}$ 的變形收縮。

命題（可縮空間是路徑連通的）:

令 $X$ 為可縮空間，它收縮到一個點 $x_{0}\in X$。那麼任意兩個點 $x,y\in X$ 可以透過一條經過 $x_{0}$ 的路徑連線起來。

證明：對於任意點$x\in X$，如果$r:[0,1]\times X\to X$ 是$X$ 到$\{x_{0}\}$ 的變形縮回，我們得到一條從$x$ 到$x_{0}$ 的路徑，方式如下：

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

它作為連續函式的複合是連續的，因為對映$t\mapsto (t,x)$ 是連續的，因為第一項是恆等函式，第二項是常數函式。 $\Box$

定義（零同倫）:

設$f:X\to Y$ 是拓撲空間之間的連續函式。 $f$ 被稱為零同倫，當且僅當它與一個常數函式同倫，即，與一個函式$g:X\to Y$ 同倫，使得存在$y_{0}\in Y$ 使得對於所有的$x\in X$，$g(x)=y_{0}$ 都成立。

命題（可縮性的刻畫）:

設$X$ 是一個拓撲空間。下列命題等價

1. $X$ 是可縮的
2. $X$ 上的恆等對映是零同倫的
3. $X$ 與單點空間同倫等價
4. 對於任意拓撲空間$Z$，任意連續函式$f:Z\to X$ 是零同倫的
5. 對於每個拓撲空間 $Z$，任何兩個連續函式 $f:Z\to X$ 和 $g:Z\to X$ 是同倫的。

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

$\gamma :=\rho _{0}*{\overline {\rho _{1}}}$，其中 $\rho _{0}(t):=H(t,z_{0})$ 和 $\rho _{1}(t):=H(t,z_{1})$。

最後，如果從任何空間到 $X$ 的任何兩個連續函式是同倫的，我們可以選擇 $Z:=X$，並將其中一個對映選擇為任何常數對映，另一個對映選擇為恆等對映；這兩個對映之間的同倫就產生了所需的形變收縮。 $\Box$

定義（同倫擴張性質）:

令 $X$ 為拓撲空間，$A\subseteq X$ 為子集。當且僅當對於任何拓撲空間 $Y$ 和任何連續函式 $f:X\to Y$，使得 $f|_{A}$ 與另一個對映 $H_{1}$ 透過同倫 $H$ 同倫，那麼存在同倫 ${\tilde {H}}$ 使得 ${\tilde {H}}_{0}=f$ 且 ${\tilde {H}}|_{[0,1]\times A}=H$ 。

命題（透過縮回表徵同倫擴張性質）:

令 $X$ 為拓撲空間，$A\subseteq X$ 為子集。$(X,A)$ 具有同倫擴張性質當且僅當存在從 $[0,1]\times X$ 到 $\{0\}\times X\cup [0,1]\times A$ 的縮回。

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

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

並聲稱它是連續的。請注意，${\tilde {H}}$ 是 $r$ （它是連續的）與對映的合成

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

因此，只需證明 $\eta$ 是連續的。為此，只需證明只要 $U\subseteq Y$ 是一個集合，使得 $U\cap \{0\}\times X$ 和 $U\cap [0,1]\times A$ 是開集，那麼 $U$ 本身就是開集；事實上，如果是這種情況，那麼只要 $V$ 是 $Z$ 的一個開子集，我們將有

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

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

對於每個 $n\in \mathbb {N}$，我們將 $O_{n}$ 定義為 $X$ 的最大開子集，使得 $[0,1/n)\times (O_{n}\cap A)\subseteq U$；由於具有此屬性的集合在並運算下是封閉的，因此存在這樣一個最大集合，我們取所有具有此屬性的集合的並集。然後定義

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

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

因此，選擇 $n\in \mathbb {N}$ 使得 $x\in O_{n}$； 那麼鄰域 $[0,1/n)\times O_{n}\cap Y$ 將作為 $(0,x)$ 的開鄰域，包含在 $U$ 內，使用子空間和乘積拓撲的定義。 $\Box$