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92 lines
3.0 KiB
Markdown
92 lines
3.0 KiB
Markdown
# Math4202 Topology II (Lecture 30)
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## Algebraic Topology
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We skipped a few chapters about Jordan curve theorem, which will be your final project soon. LOL, I will embedded the link once I'm done.
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### Seifert-Van Kampen Theorem
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#### The Seifert-Van Kampen Theorem
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Let $X=U\cup V$ be a union of two open subspaces. Suppose that $U\cap V$, $U,V$ are path connected. Fix $x_0\in U\cap V$.
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Let $H$ be a group (arbitrary). And now we assume $\phi_1,\phi_2$ be a group homomorphism, and $\phi_1:\pi_1(U,x_0)\to H$, and $\phi_2:\pi_1(V,x_0)\to H$.
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Let $i_1,i_2,j_1,j_2,i_{12}$ be group homomorphism induced by the inclusion maps.
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Assume this diagram commutes.
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$$
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\phi_1\circ i_1=\phi_2\circ i_2
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$$
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There is a group homomorphism $\Phi:\pi_1(X,x_0)\to H$ making the diagram commute. $\Phi\circ j_1=\phi_1$ and $\Phi\circ j_2=\phi_2$.
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We may change the base point using conjugations.
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<details>
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<summary>Side notes about free product of two groups</summary>
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Consider arbitrary group $G_1,G_2$, then $G_1\times G_2$ is a group.
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Note that the inclusion map $i_1:G_1\to G_1\times G_2$ is a group homomorphism and the inclusion map $i_2:G_2\to G_1\times G_2$ is a group homomorphism. The image of them commutes since $(e,g_2)(g_1,e)=(g_1,g_2)=(g_1,e)(e,g_2)$.
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#### The universal property
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Then we want to have a group $G$ such that for all group homomorphism $\phi:G_1\to H$ and $G_2\to H$, such that there always exists a map $\Phi: G\to H$ such that:
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- $\Phi\circ i_1=\phi_1$
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- $\Phi\circ i_2=\phi_2$
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#### How to construct the free group?
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We consider
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$$
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G_1*G_2=S=\{g_1h_1g_2h_2:g_1,g_2\in G_1,h_1,h_2\in G_2\}/\sim
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$$
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And we set $g_ie_{G_2}g_{i+1}\sim g_ig_{i+1}$ for $g_i\in G_1$ and $g_{i+1}\in G_2$.
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And $h_je_{G_1}h_{j+1}\sim h_jh_{j+1}$ for $h_j\in G_2$ and $h_{j+1}\in G_1$.
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And we define the group operation
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$$
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(g_1 h_1\cdots g_k h_k)*(h_1' g_1'\cdots h_l' g_l')=g_1 h_1\cdots g_k h_k g_1' h_2'\cdots h_l' g_l'
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$$
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And the inverse is defined
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$$
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(g_1 h_1\cdots g_k h_k)^{-1}=h_k^{-1} g_k^{-1}\cdots h_1^{-1} g_1^{-1}
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$$
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And $G=S$ is a well-defined group.
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The homeomorphism $G\to H$ is defined as
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$$
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\Phi((g_1 h_1\cdots g_k h_k))=\phi_1(g_1)\circ \phi_2(h_1)\circ \cdots \circ \phi_1(g_k)\circ \phi_2(h_k)
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$$
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Note $\circ$ is the group operation in $H$.
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> Group with such universal property is unique, so we don't need to worry for that too much.
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</details>
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Back to the Seifert-Van Kampen Theorem:
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Let $H=\pi_1(U,x_0)* \pi_1(V,x_0)$.
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Let $N$ be the **least normal subgroup** in the free product $H$, containing $i_1(g)i_2(g)^{-1}$, $\forall g\in \pi_1(U\cap V,x_0)$.
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Note $i_1(g)\in \pi_1(U,x_0)$ and $i_2(g)\in \pi_1(V,x_0)$. You may think of them as $G_1,G_2$ in the free group descriptions.
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#### Seifert-Van Kampen Theorem (classical version)
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There is an isomorphism between $\pi_1(U,x_0)* \pi_1(V,x_0)/N$ and $\pi_1(U\cup V,x_0)$.
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