環面 球面 理解基本群概念的一種簡單方法是從一個具體的例子開始。讓我們考慮二維球面 S 2 {\displaystyle S^{2}}
讓我們開始考慮環面上的兩種型別的迴路（起點和終點相同的路徑）。似乎繞環面“臂”的路徑與“區域性”簡單迴路有本質區別：一個不能變形為另一個。另一方面，在球面上，似乎所有的迴路都可以變形為任何其他迴路。這兩個空間中的“迴路型別”集合是不同的：環面似乎比球面表面具有更豐富的“迴路型別”集合。這種型別的思路構成了基本群定義的基礎，並解釋了不同型別的拓撲空間之間的本質區別。基本群使這個想法在數學上變得嚴格。
定義： 令 X {\displaystyle X} p {\displaystyle p} q {\displaystyle q} X {\displaystyle X} f : [ 0 , 1 ] → X {\displaystyle f:[0,1]\rightarrow X} g : [ 0 , 1 ] → X {\displaystyle g:[0,1]\rightarrow X} H : [ 0 , 1 ] 2 → X {\displaystyle H:[0,1]^{2}\rightarrow X} a ∈ [ 0 , 1 ] {\displaystyle a\in [0,1]} H ( x , a ) {\displaystyle H(x,a)} p {\displaystyle p} q {\displaystyle q}
定義： 定義路徑的組合 f 1 {\displaystyle f_{1}} x {\displaystyle x} y {\displaystyle y} f 2 {\displaystyle f_{2}} y {\displaystyle y} z {\displaystyle z}
f ( x ) = { f 1 ( 2 x ) if x ∈ [ 0 , 1 2 ] f 2 ( 2 x − 1 ) if x ∈ [ 1 2 , 1 ] {\displaystyle f(x)=\left\{{\begin{array}{ll}f_{1}(2x)&{\text{if }}x\in [0,{\frac {1}{2}}]\\f_{2}(2x-1)&{\text{if }}x\in [{\frac {1}{2}},1]\\\end{array}}\right.}
我們將用 f ( x ) ∗ g ( x ) {\displaystyle f(x)*g(x)} f ( x ) {\displaystyle f(x)} g ( x ) {\displaystyle g(x)}
定義： 令 f : [ 0 , 1 ] → X {\displaystyle f:[0,1]\rightarrow X} f − 1 ( x ) = f ( 1 − x ) {\displaystyle f^{-1}(x)=f(1-x)}
定義： 令 X {\displaystyle X} p {\displaystyle p} X {\displaystyle X} C p {\displaystyle C_{p}} f : [ 0 , 1 ] → X {\displaystyle f:[0,1]\rightarrow X} f ( x ) = p {\displaystyle f(x)=p}
現在考慮路徑的等價類集合。定義兩個等價類的合成是任何兩個路徑的合成的等價類。定義一個等價類的逆是該等價類中任何路徑的逆的等價類。定義 [ C p ] {\displaystyle [C_{p}]} C p {\displaystyle C_{p}}
我們可以很容易地檢查這些操作是否定義良好。
現在，在基本群中，我們將使用迴圈。因此，我們定義迴圈的等價性、合成和逆與路徑的定義相同，等價類的合成和逆也相同。
定義： 基點 x 0 {\displaystyle x_{0}} X {\displaystyle X} x 0 {\displaystyle x_{0}} 基本群 。
為了證明這是一個群，我們需要證明
1) 結合律： [ α ] ∗ ( [ β ] ∗ [ γ ] ) = ( [ α ] ∗ [ β ] ) ∗ [ γ ] {\displaystyle [\alpha ]*([\beta ]*[\gamma ])=([\alpha ]*[\beta ])*[\gamma ]}
2) 單位元： [ α ] ∗ [ 1 ] = [ 1 ] ∗ [ α ] = [ α ] {\displaystyle [\alpha ]*[1]=[1]*[\alpha ]=[\alpha ]}
3) 逆元： [ α ] ∗ [ α ¯ ] = [ α ¯ ] ∗ [ α ] = [ 1 ] {\displaystyle [\alpha ]*[{\overline {\alpha }}]=[{\overline {\alpha }}]*[\alpha ]=[1]}
1) 很明顯，當您從 a {\displaystyle a} b {\displaystyle b} b {\displaystyle b} c {\displaystyle c} c {\displaystyle c} d {\displaystyle d} a {\displaystyle a} b {\displaystyle b} b {\displaystyle b} d {\displaystyle d} a {\displaystyle a} c {\displaystyle c} c {\displaystyle c} d {\displaystyle d}
事實上，關於 [ α ] ∗ ( [ β ] ∗ [ γ ] ) {\displaystyle [\alpha ]*([\beta ]*[\gamma ])} ( [ α ] ∗ [ β ] ) ∗ [ γ ] {\displaystyle ([\alpha ]*[\beta ])*[\gamma ]}
F ( t , s ) = { α ( 4 t s + 1 ) , if 0 ≤ t ≤ s + 1 4 β ( 4 t − s − 1 ) , if s + 1 4 ≤ t ≤ s + 2 4 γ ( 4 t − s − 2 2 − s ) , if s + 2 4 ≤ t ≤ 1 {\displaystyle F(t,s)={\begin{cases}\alpha ({\frac {4t}{s+1}}),&{\mbox{if }}0\leq t\leq {\frac {s+1}{4}}\\\beta (4t-s-1),&{\mbox{if }}{\frac {s+1}{4}}\leq t\leq {\frac {s+2}{4}}\\\gamma ({\frac {4t-s-2}{2-s}}),&{\mbox{if }}{\frac {s+2}{4}}\leq t\leq 1\end{cases}}} .
2) [ C x 0 ] {\displaystyle [C_{x_{0}}]}
3) 之前定義的等價關係的逆元作為群中的逆元。兩個路徑的組合 f ( x ) {\displaystyle f(x)} f − 1 ( x ) {\displaystyle f^{-1}(x)}
h ( x , y ) = f ( x y ) ∗ f − 1 ( x y ) {\displaystyle h(x,y)=f(xy)*f^{-1}(xy)}
◻ {\displaystyle \Box }
我們現在有了我們的基本群，但瞭解基本群如何依賴於基點會很有趣，因為根據我們的定義，基本群依賴於基點。然而，由於在任何路徑連通的拓撲空間中，其所有基本群都同構的非常重要的定理，我們能夠對任何路徑連通的拓撲空間談論拓撲空間的基本群 。
讓我們取同一個路徑連通分量的 x 0 , x 1 ∈ X {\displaystyle x_{0},x_{1}\in X} X {\displaystyle X} π 1 ( X , x 0 ) {\displaystyle \pi _{1}(X,x_{0})} π 1 ( X , x 1 ) {\displaystyle \pi _{1}(X,x_{1})} h : [ 0 , 1 ] → X {\displaystyle h:\left[0,1\right]\to X} x 0 {\displaystyle x_{0}} x 1 {\displaystyle x_{1}} h ¯ ( s ) = 1 − s {\displaystyle {\overline {h}}(s)=1-s} x 1 {\displaystyle x_{1}} x 0 {\displaystyle x_{0}} β h [ f ] = [ h f h ¯ ] {\displaystyle \beta _{h}[f]=[hf{\overline {h}}]} β h : π 1 ( X , x 1 ) → π 1 ( X , x 0 ) {\displaystyle \beta _{h}:\pi _{1}(X,x_{1})\to \pi _{1}(X,x_{0})} X {\displaystyle X} π 1 ( X , x 0 ) {\displaystyle \pi _{1}(X,x_{0})} x 0 {\displaystyle x_{0}}
當一個拓撲空間的所有基本群都同構時，記號 π 1 ( X , x 0 ) {\displaystyle \pi _{1}(X,x_{0})} π 1 ( X ) {\displaystyle \pi _{1}(X)}
定義： 如果一個拓撲空間 X {\displaystyle X}
S 1 {\displaystyle S^{1}} [ 編輯 | 編輯原始碼 ] 本節致力於計算圓周 S 1 {\displaystyle S^{1}}
很容易想象繞圓周的一圈路徑與平凡路徑不同倫。同樣很容易想象繞兩圈的路徑與繞一圈的路徑不同倫。直觀的感覺似乎是 S 1 {\displaystyle S^{1}} π 1 ( S 1 ) {\displaystyle \pi _{1}(S^{1})}
我們定義 p : R → S 1 ⊂ C , p ( t ) = e 2 i π t {\displaystyle p:\mathbb {R} \to S^{1}\subset \mathbb {C} ,p(t)=e^{2i\pi t}}
引理 1： 設 f : [ 0 , 1 ] → S 1 ⊂ C {\displaystyle f:\left[0,1\right]\to S^{1}\subset \mathbb {C} } f ¯ : [ 0 , 1 ] → R {\displaystyle {\overline {f}}:\left[0,1\right]\to \mathbb {R} } p ∘ f ¯ = f {\displaystyle p\circ {\overline {f}}=f} f ( 0 ) = x 0 ∧ z 0 ∈ p − 1 ( x 0 ) {\displaystyle f(0)=x_{0}\land z_{0}\in p^{-1}(x_{0})} f ¯ {\displaystyle {\overline {f}}} f {\displaystyle f}
引理 2： 設 F : [ 0 , 1 ] × [ 0 , 1 ] → S 1 ⊂ C {\displaystyle F:\left[0,1\right]\times \left[0,1\right]\to S^{1}\subset \mathbb {C} } x 0 {\displaystyle x_{0}} z 0 ∈ p − 1 ( x 0 ) {\displaystyle z_{0}\in p^{-1}(x_{0})} F ¯ : [ 0 , 1 ] × [ 0 , 1 ] → R {\displaystyle {\overline {F}}:\left[0,1\right]\times \left[0,1\right]\to \mathbb {R} } z 0 {\displaystyle z_{0}} p ∘ F ¯ = F {\displaystyle p\circ {\overline {F}}=F}
注意： 這些引理保證了同倫的環路有同倫的提升。
更多資訊請參閱 維基百科 。
定理： π 1 ( S 1 , ( 1 , 0 ) ) ≃ Z {\displaystyle \pi _{1}(S^{1},(1,0))\simeq \mathbb {Z} }
證明： 設 α {\displaystyle \alpha } ( 1 , 0 ) {\displaystyle (1,0)} [ α ] ∈ π 1 ( S 1 , ( 1 , 0 ) ) {\displaystyle [\alpha ]\in \pi _{1}(S^{1},(1,0))} z 0 = 0 {\displaystyle z_{0}=0}
π 1 ( S 1 , ( 1 , 0 ) ) → Z v ( [ α ] ) = α ¯ ( 1 ) . {\displaystyle \pi _{1}(S^{1},(1,0))\to \mathbb {Z} \;\;\;v([\alpha ])={\overline {\alpha }}(1).} 此應用的良好定義源於在 [ 0 , 1 ] {\displaystyle \left[0,1\right]} S 1 {\displaystyle S^{1}} p ( α ¯ ( 1 ) ) = α ( 1 ) = x 0 {\displaystyle p({\overline {\alpha }}(1))=\alpha (1)=x_{0}} α ¯ ( 1 ) = k {\displaystyle {\overline {\alpha }}(1)=k} k ∈ Z {\displaystyle k\in \mathbb {Z} } v ( [ α ] ) ∈ Z {\displaystyle v([\alpha ])\in \mathbb {Z} }
1) v {\displaystyle v} N ∈ Z {\displaystyle N\in \mathbb {Z} } α N ( t ) = e 2 i π N t {\displaystyle \alpha _{N}(t)=e^{2i\pi Nt}} α ¯ N ( t ) = N t {\displaystyle {\overline {\alpha }}_{N}(t)=Nt} v ( [ α N ] ) = N {\displaystyle v([\alpha _{N}])=N}
2) v {\displaystyle v} v ( [ α ] ) = v ( [ β ] ) {\displaystyle v([\alpha ])=v([\beta ])} ( α ¯ ( 1 ) = β ¯ ( 1 ) ) ∧ ( α ¯ ( 0 ) = β ¯ ( 0 ) = 0 ) {\displaystyle ({\overline {\alpha }}(1)={\overline {\beta }}(1))\land ({\overline {\alpha }}(0)={\overline {\beta }}(0)=0)} F ( t , s ) = p ( ( 1 − s ) α ¯ ( t ) + s β ¯ ( t ) ) {\displaystyle F(t,s)=p((1-s){\overline {\alpha }}(t)+s{\overline {\beta }}(t))} α {\displaystyle \alpha } β {\displaystyle \beta } [ α ] = [ β ] {\displaystyle [\alpha ]=[\beta ]}
3) v {\displaystyle v} v ( [ α ] ∗ [ β ] ) = v ( [ α ] + v ( [ β ] {\displaystyle v([\alpha ]*[\beta ])=v([\alpha ]+v([\beta ]}
γ ( t ) = { α ¯ ( 2 t ) , if 0 ≤ t ≤ 1 2 β ¯ ( 2 t − 1 ) + α ¯ ( 1 ) , if 1 2 ≤ t ≤ 1 {\displaystyle \gamma (t)={\begin{cases}{\overline {\alpha }}(2t),&{\mbox{if }}0\leq t\leq {\frac {1}{2}}\\{\overline {\beta }}(2t-1)+{\overline {\alpha }}(1),&{\mbox{if }}{\frac {1}{2}}\leq t\leq 1\end{cases}}} α ¯ ( 1 ) {\displaystyle {\overline {\alpha }}(1)} p ∘ γ = α ∗ β {\displaystyle p\circ \gamma =\alpha *\beta } γ = α ∗ β ¯ {\displaystyle \gamma ={\overline {\alpha *\beta }}}
v ( [ α ] ∗ [ β ] ) = v ( [ α ∗ β ] ) = γ ( 1 ) = α ¯ ( 1 ) + β ¯ ( 1 ) = v ( [ α ] ) + v ( [ β ] ) {\displaystyle v([\alpha ]*[\beta ])=v([\alpha *\beta ])=\gamma (1)={\overline {\alpha }}(1)+{\overline {\beta }}(1)=v([\alpha ])+v([\beta ])} 我們可以注意到，所有迴圈都與 α N ( t ) = e 2 i π N t {\displaystyle \alpha _{N}(t)=e^{2i\pi Nt}} N ∈ Z {\displaystyle N\in \mathbb {Z} }
我們可以透過以下方案來理解這個證明：
◻ {\displaystyle \Box }
研究基本群的一個最有用的工具是覆蓋空間 。直觀地說，給定空間 X {\displaystyle X} X {\displaystyle X} X {\displaystyle X}
本節將正式定義覆蓋空間，陳述覆蓋空間的重要提升定理 ，然後說明它們對基本群的影響。
定義： 假設 X {\displaystyle X} p : Z → X {\displaystyle p:Z\rightarrow X} x ∈ X {\displaystyle x\in X} U ∋ x {\displaystyle U\ni x}
(i) p − 1 ( U ) {\displaystyle p^{-1}(U)} Z {\displaystyle Z} ∐ α ∈ A V α {\displaystyle \coprod _{\alpha \in A}V_{\alpha }}
(ii) p {\displaystyle p} V α {\displaystyle V_{\alpha }} V α {\displaystyle V_{\alpha }}
p {\displaystyle p} 覆蓋對映 。
示例： 事實上，我們已經看到了覆蓋空間的一個例子。在計算 π 1 ( S 1 ) {\displaystyle \pi _{1}(S^{1})} R {\displaystyle \mathbb {R} } S 1 {\displaystyle S^{1}} p : R → S 1 , t ↦ e 2 π i t {\displaystyle p:\mathbb {R} \rightarrow S^{1},t\mapsto e^{2\pi it}} e 2 π i t 1 = e 2 π i t 2 {\displaystyle e^{2\pi it_{1}}=e^{2\pi it_{2}}} t 1 − t 2 {\displaystyle t_{1}-t_{2}} x ∈ S 1 {\displaystyle x\in S^{1}} U x = S 1 − { − x } {\displaystyle U_{x}=S^{1}-\{-x\}} x {\displaystyle x} A r g ( x ) = t x {\displaystyle Arg(x)=t_{x}} p − 1 ( U ) = R − { t x + Z } {\displaystyle p^{-1}(U)=\mathbb {R} -\{t_{x}+\mathbb {Z} \}} U {\displaystyle U} t x + n , n ∈ Z {\displaystyle t_{x}+n,n\in \mathbb {Z} }
很明顯（畫個圖！），這個集合是子區間的互不相交的並集，可以檢驗指數函式將每個子區間同胚對映到 U x {\displaystyle U_{x}} ◻ {\displaystyle \square }
現在我們來介紹一個定理，它乍一看可能有點深奧，但實際上它使我們能夠在覆蓋空間上做很多事情。
定理（同倫提升）： 假設 p : Z → X {\displaystyle p:Z\rightarrow X} X {\displaystyle X} f : I n → X {\displaystyle f:I^{n}\rightarrow X} n {\displaystyle n} X {\displaystyle X} F : I n + 1 → X {\displaystyle F:I^{n+1}\rightarrow X} f {\displaystyle f} f ′ : I n → X {\displaystyle f':I^{n}\rightarrow X} ϕ : I n → Z {\displaystyle \phi :I^{n}\rightarrow Z} p ⋅ ϕ = f {\displaystyle p\cdot \phi =f} Φ : I n + 1 → Z {\displaystyle \Phi :I^{n+1}\rightarrow Z}
(i) Φ | I n = ϕ {\displaystyle \Phi |_{I^{n}}=\phi }
(ii) p ⋅ Φ = F {\displaystyle p\cdot \Phi =F} ◻ {\displaystyle \square }
n = 0 {\displaystyle n=0} I n {\displaystyle I^{n}} f : I n → X {\displaystyle f:I^{n}\rightarrow X} X {\displaystyle X} f {\displaystyle f} x 0 ∈ X {\displaystyle x_{0}\in X} f {\displaystyle f} F : I → X {\displaystyle F:I\rightarrow X} F ( 0 ) = x 0 {\displaystyle F(0)=x_{0}} X {\displaystyle X} x 0 {\displaystyle x_{0}} p : Z → X {\displaystyle p:Z\rightarrow X}
它說（檢查一下！），如果 z 0 ∈ Z {\displaystyle z_{0}\in Z} p ( z 0 ) = x 0 {\displaystyle p(z_{0})=x_{0}} γ {\displaystyle \gamma } X {\displaystyle X} x 0 {\displaystyle x_{0}} Z {\displaystyle Z} 唯一 的路徑 γ ′ {\displaystyle \gamma '} z 0 {\displaystyle z_{0}} p γ ˙ ′ = γ {\displaystyle p{\dot {\gamma }}'=\gamma } X {\displaystyle X} 唯一 的提升到 Z {\displaystyle Z}
回顧一下，這個結果——有時被稱為路徑提升定理 ——並不那麼令人驚訝。將覆蓋空間想象成基空間 X {\displaystyle X} U ⊂ X {\displaystyle U\subset X} Z {\displaystyle Z} γ {\displaystyle \gamma } U {\displaystyle U} V α {\displaystyle V_{\alpha }} V α {\displaystyle V_{\alpha }} p {\displaystyle p} γ {\displaystyle \gamma } V α {\displaystyle V_{\alpha }} x 0 {\displaystyle x_{0}} γ {\displaystyle \gamma } U {\displaystyle U}
現在我們來談談覆蓋空間和基本群之間的聯絡，這是一個非常重要的聯絡。
定理： 給定一個覆蓋空間 p : Z → X {\displaystyle p:Z\rightarrow X} p {\displaystyle p} p ∗ : π 1 ( Z ) → π 1 ( X ) {\displaystyle p_{*}:\pi _{1}(Z)\rightarrow \pi _{1}(X)}
證明（概述）： 首先，考慮一條路徑 γ {\displaystyle \gamma } Z {\displaystyle Z} γ : I → Z {\displaystyle \gamma :I\rightarrow Z} p {\displaystyle p} p ⋅ γ {\displaystyle p\cdot \gamma } X {\displaystyle X}
p ′ : {\displaystyle p':} Z → {\displaystyle Z\rightarrow } X {\displaystyle X} 我們想證明這可以用來定義一個對映
p ∗ : {\displaystyle p*:} Z {\displaystyle Z} z 0 → {\displaystyle z_{0}\rightarrow } → {\displaystyle \rightarrow } X {\displaystyle X} x 0 {\displaystyle x_{0}} 這聽起來很複雜，但實際上一點也不復雜：想法是，給定一個同倫 H : I × I → Z {\displaystyle H:I\times I\rightarrow Z} γ 1 {\displaystyle \gamma _{1}} γ 2 {\displaystyle \gamma _{2}} Z {\displaystyle Z} p ⋅ H {\displaystyle p\cdot H} X {\displaystyle X} p ′ ( γ 1 ) {\displaystyle p'(\gamma _{1})} p ′ ( γ 2 ) {\displaystyle p'(\gamma _{2})} z 0 {\displaystyle z_{0}} p ( z 0 ) {\displaystyle p(z_{0})}
所以，我們得到了我們想要的對映 p ∗ {\displaystyle p*} π 1 ( Z , z 0 ) {\displaystyle \pi _{1}(Z,z_{0})} π 1 ( X , p ( z 0 ) ) . {\displaystyle \pi _{1}(X,p(z_{0})).}
(a) 很容易。為了證明它，選擇 π 1 ( Z , z 0 ) {\displaystyle \pi _{1}(Z,z_{0})} X {\displaystyle X} p ′ {\displaystyle p'}
(b) is more tricky. To prove it, we must show that the kernel of the homomorphism p ∗ {\displaystyle p*} p i 1 ( Z , z 0 ) {\displaystyle pi_{1}(Z,z_{0})} γ {\displaystyle \gamma } [ γ ] {\displaystyle [\gamma ]} p ∗ ( [ γ ] ) {\displaystyle p*([\gamma ])} π 1 ( X , p ( z 0 ) ) {\displaystyle \pi _{1}(X,p(z_{0}))} p ∗ {\displaystyle p*} p ′ ( γ ) {\displaystyle p'(\gamma )} X {\displaystyle X} p ( z 0 ) {\displaystyle p(z_{0})} F : I × I → X {\displaystyle F:I\times I\rightarrow X} F {\displaystyle F} F ′ {\displaystyle F'} Z {\displaystyle Z} γ {\displaystyle \gamma } z 0 {\displaystyle z_{0}} F ′ {\displaystyle F'} [ γ ] {\displaystyle [\gamma ]} π 1 ( Z , z 0 ) {\displaystyle \pi _{1}(Z,z_{0})} p ∗ {\displaystyle p*} π 1 ( Z , z 0 ) {\displaystyle \pi _{1}(Z,z_{0})} p ∗ {\displaystyle p*} ◻ {\displaystyle \square }
讓我們花點時間思考一下這個結果的意義。群的單射同態 G → H {\displaystyle G\rightarrow H} G H ⊂ H {\displaystyle G_{H}\subset H} X {\displaystyle X} Z {\displaystyle Z} π 1 ( Z ) {\displaystyle \pi _{1}(Z)} π 1 ( X ) {\displaystyle \pi _{1}(X)} X {\displaystyle X} S 1 {\displaystyle S^{1}} R {\displaystyle \mathbb {R} } S 1 × S 1 {\displaystyle S^{1}\times S^{1}} S 2 {\displaystyle S^{2}} Z 2 {\displaystyle \mathbb {Z} ^{2}}
我認為，後一個例子特別好，因為它表明，透過檢視空間的代數不變式（在本例中，是基團），而不是我們對空間本身的幾何“心理影像”，可以大大簡化關於空間之間特定型別對映的存在或形式的論證。 你能給出簡單的幾何論證來證明，不可能將 S 2 {\displaystyle S^{2}} S 3 {\displaystyle S^{3}} S n {\displaystyle S^{n}}
示例： 現在讓我們來看一個具體的覆蓋空間，並看看我們上面提到的同態對映 p ∗ {\displaystyle p*} 是什麼 。把圓 S 1 {\displaystyle S^{1}} s 1 = { z ∈ C : | z | = 1 } {\displaystyle s^{1}=\{z\in \mathbb {C} :|z|=1\}} p : S 1 → S 1 {\displaystyle p:S^{1}\rightarrow S^{1}} p ( z ) = z 2 {\displaystyle p(z)=z^{2}}
我認為 p {\displaystyle p} e i θ {\displaystyle e^{i\theta }} S 1 {\displaystyle S^{1}} 0 ≤ θ < 2 π {\displaystyle 0\leq \theta <2\pi } z ′ ∈ S 1 {\displaystyle z'\in S^{1}} z {\displaystyle z} p {\displaystyle p} z {\displaystyle z} p {\displaystyle p}
所以， p {\displaystyle p} p ∗ {\displaystyle p*} p ∗ : Z → Z {\displaystyle p^{*}:\mathbb {Z} \rightarrow \mathbb {Z} } γ {\displaystyle \gamma } S 1 {\displaystyle S^{1}} 1 ∈ Z {\displaystyle 1\in \mathbb {Z} } π 1 ( S 1 ) ≅ Z {\displaystyle \pi _{1}(S^{1})\cong \mathbb {Z} } p ∗ [ γ ] {\displaystyle p*[\gamma ]} p ⋅ γ {\displaystyle p\cdot \gamma } S 1 {\displaystyle S^{1}} p ∗ {\displaystyle p*} p ⋅ γ {\displaystyle p\cdot \gamma } S 1 {\displaystyle S^{1}} [ γ ] ∗ [ γ ] {\displaystyle [\gamma ]*[\gamma ]} p ∗ ( [ γ ] ) = [ γ ] ∗ [ γ ] {\displaystyle p*([\gamma ])=[\gamma ]*[\gamma ]} π 1 ( S 1 ) {\displaystyle \pi _{1}(S^{1})} p ∗ ( 1 ) = 2 {\displaystyle p*(1)=2} p ∗ : Z → Z {\displaystyle p^{*}:\mathbb {Z} \rightarrow \mathbb {Z} }
當然，引入覆蓋空間之類的概念很好，但除非我們的新概念在某種程度上被證明有用，否則不會有什麼作用。覆蓋空間確實在很多方面被證明有用，但希望以下示例足以說明這一點。
定理（Nielsen-Schreier）： 自由群的任何子群都是自由的。
證明（概要）： 查閱維基百科以獲得對“自由群”的嚴格定義：大致來說，它是一個非平凡元素組合不等於單位元的群。現在，證明策略如下
1) 給定一個自由群 F {\displaystyle F} X {\displaystyle X} π 1 ( X ) = F {\displaystyle \pi _{1}(X)=F} 圖 是一個拓撲空間，它由一組離散點組成，這些點連線著一族線段。請再次參考維基百科以瞭解嚴格定義。)
2) 證明對於一個空間 X {\displaystyle X} H < π 1 ( X ) {\displaystyle H<\pi _{1}(X)} X {\displaystyle X} Z {\displaystyle Z} π 1 ( Z ) = H {\displaystyle \pi _{1}(Z)=H}
3) 證明任何圖的覆蓋空間本身也是一個圖。
4) 證明圖的基本群是一個自由群。
定理：設 f 是一個在複數 C {\displaystyle \mathbb {C} } C {\displaystyle \mathbb {C} } C {\displaystyle \mathbb {C} }
Suppose that p ∈ C [ z ] {\displaystyle p\in \mathbb {C} [z]} C {\displaystyle \mathbb {C} } p {\displaystyle p} p ( z ) = z n + ∑ i = 1 n a i z n − i {\displaystyle p(z)=z^{n}+\sum _{i=1}^{n}a_{i}z^{n-i}} q ∈ C [ z ] {\displaystyle q\in \mathbb {C} [z]} f q : [ 0 , ∞ ) → Ω ( S 1 , 1 ) {\displaystyle f_{q}:[0,\infty )\rightarrow \Omega (S^{1},1)} f q ( r ) ( s ) = q ( r exp ( 2 π i s ) ) / q ( r ) | q ( r exp ( 2 π ) i s ) / q ( r ) | . {\displaystyle {\begin{aligned}f_{q}(r)(s)={\frac {q(r\exp(2\pi is))/q(r)}{\vert q(r\exp(2\pi )is)/q(r)\vert }}.\end{aligned}}} f q ( r ) {\displaystyle f_{q}(r)} S 1 {\displaystyle S^{1}} r {\displaystyle r} r ′ ∈ [ 0 , ∞ ) {\displaystyle r^{\prime }\in [0,\infty )} H r ′ : I × I → S 1 {\displaystyle H_{r^{\prime }}:I\times I\rightarrow S^{1}} f p ( 0 ) {\displaystyle f_{p}(0)} f p ( r ′ ) {\displaystyle f_{p}(r^{\prime })} H r ′ ( s , t ) = f p ( t r ) ( s ) {\displaystyle H_{r^{\prime }}(s,t)=f_{p}(tr)(s)} f p ( 0 ) {\displaystyle f_{p}(0)} 1 ∈ S 1 {\displaystyle 1\in S^{1}} f p ( r ) {\displaystyle f_{p}(r)} r {\displaystyle r}
現在我們證明，對於 r {\displaystyle r} f p ( r ) {\displaystyle f_{p}(r)} ω n : I → S 1 {\displaystyle \omega _{n}:I\rightarrow S^{1}} ω n ( s ) = exp ( 2 π i n s ) {\displaystyle \omega _{n}(s)=\exp(2\pi ins)} n {\displaystyle n} p {\displaystyle p} ω := ω 1 {\displaystyle \omega :=\omega _{1}} π 1 ( S 1 , 1 ) {\displaystyle \pi _{1}(S^{1},1)} ω n {\displaystyle \omega _{n}} ω n {\displaystyle \omega ^{n}} f p ( r ) {\displaystyle f_{p}(r)} n = 0 {\displaystyle n=0} p {\displaystyle p}
為此，固定 r 0 > max { 1 , ∑ i = 1 n | a i | } {\displaystyle r_{0}>\max \lbrace 1,\sum _{i=1}^{n}\vert a_{i}\vert \rbrace } C {\displaystyle C} r 0 {\displaystyle r_{0}} z ∈ C {\displaystyle z\in C} t ∈ [ 0 , 1 ] {\displaystyle t\in [0,1]} | z n | > | z n − 1 | ∑ i = 1 n | a i | > ∑ i = 1 n | a i z n − i | ≥ | ∑ i = 1 n a i z n − i | ≥ t | ∑ i = 1 n a i z n − i | . {\displaystyle {\begin{aligned}\vert z^{n}\vert >&\vert z^{n-1}\vert \sum _{i=1}^{n}\vert a_{i}\vert \\>&\sum _{i=1}^{n}\vert a_{i}z^{n-i}\vert \\\geq &\left\vert \sum _{i=1}^{n}a_{i}z^{n-i}\right\vert \\\geq &t\left\vert \sum _{i=1}^{n}a_{i}z^{n-i}\right\vert .\\\end{aligned}}}
這意味著對於所有 t ∈ [ 0 , 1 ] {\displaystyle t\in [0,1]} p t ( z ) := z n + t ∑ i = 1 n a i z n − i {\displaystyle p_{t}(z):=z^{n}+t\sum _{i=1}^{n}a_{i}z^{n-i}} C {\displaystyle C} | z n | = | t | | ∑ i = 1 n a i z n − i | {\displaystyle \vert z^{n}\vert =\vert t\vert \left\vert \sum _{i=1}^{n}a_{i}z^{n-i}\right\vert } J : I × I → S 1 {\displaystyle J:I\times I\rightarrow S^{1}} J ( s , t ) = f p t ( s ) {\displaystyle J(s,t)=f_{p_{t}}(s)} J {\displaystyle J} ω n {\displaystyle \omega _{n}} f p ( r 0 ) {\displaystyle f_{p}(r_{0})} n = 0 {\displaystyle n=0}
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