Math4202 Topology II (Lecture 32)

Consider the following three spaces:

$$\pi_1(U,x_0)* \pi_1(V,x_0)$$

there is a quiotient quotient map $q:\pi_1(U,x_0)* \pi_1(V,x_0)\to \pi_1(U\cup V,x_0)=\to \pi_1(U,x_0)* \pi_1(V,x_0)/N$ $q:\pi_1(U,x_0)* \pi_1(V,x_0)\to \pi_1(U,x_0)* \pi_1(V,x_0)/N$ (the quotient map lands in the quotient group, not in $\pi_1(U\cup V,x_0)$).

And $j_1\circ i_1=j_2\circ i_2$.

Motivation: This commutativity of the inclusion-induced square is the whole reason $j$ kills every generator of $N$ below; without it the descent to the quotient would fail.

Let $k\in \pi_1(U,x_0)* \pi_1(V,x_0)$, we need to show that

$$\forall k\in \pi_1(U,x_0)* \pi_1(V,x_0),\quad g\in \pi_1(U\cap V,x_0)$$

~~$$ \begin{aligned} j(ki_1(k)^{-1}(g)i_2(g)k^{-1})&=j(k)j_1(i_1(g)^{-1})j_2(i_2(g))j(k^{-1})\ &=e_X \end{aligned}

$$$$

\begin{aligned} j(k\cdot i_1(g)^{-1}i_2(g)\cdot k^{-1})&=j(k),j_1(i_1(g))^{-1},j_2(i_2(g)),j(k)^{-1}\ &=j(k),(j_1\circ i_1)(g)^{-1},(j_2\circ i_2)(g),j(k)^{-1}\ &=j(k),j(k)^{-1}=e_X \end{aligned}

_(the middle factor cancels because $j_1\circ i_1=j_2\circ i_2$)._ Since $\{ki_1(g)^{-1}i_2(g)k^{-1}|g,k\}$ generates $N$, _it follows that $j(N)=\{e_X\}$, so $j$ descends to a well-defined homomorphism on the quotient._ Consider $\tilde{j}:\pi_1(U,x_0)* \pi_1(V,x_0)/N\to \pi_1(U\cup V,x_0)=\pi_1(X,x_0)$. ~~$\forall [f]\in \pi_1(U,x_0)* \pi_1(V,x_0)/N$, we need to find $f\in \pi_1(U,x_0)* \pi_1(V,x_0)$ such that $[f]=[f]$.~~ _Incentive to continue: What we actually need now is an inverse $\Phi:\pi_1(X,x_0)\to\pi_1(U,x_0)*\pi_1(V,x_0)/N$ for $\tilde j$. Apply the universal property (S-vK) with target $H=\pi_1(U,x_0)*\pi_1(V,x_0)/N$ and with $\phi_1,\phi_2$ the compositions of the inclusions into the free product with $q$; these satisfy $\phi_1\circ i_1=\phi_2\circ i_2$ because $i_1(g)^{-1}i_2(g)\in N$. This yields a unique $\Phi$ such that $\Phi\circ j_1=\phi_1$ and $\Phi\circ j_2=\phi_2$._ /*Track lost*/ > It basically says that the loop generated by taking isomorphism first and then quotient is the same as the loop generated by taking the inclusion first and then homomorphism. (If that is real I guess.) _Bridge: Observe that both $\tilde j\circ\Phi$ and $\mathrm{id}_{\pi_1(X,x_0)}$ fit into the universal-property diagram with $H=\pi_1(X,x_0)$ and $\phi_i=j_i$ (for $\tilde j\circ\Phi$ this uses $\tilde j\circ\phi_i=j_i$, i.e. $\tilde j\circ q\circ(\text{incl}_i)=j_i$, which is how $\tilde j$ was constructed). Uniqueness in the universal property forces these two maps to agree._ The map $\tilde{j}\circ \Phi:\pi_1(X,x_0)\to \pi_1(X,x_0)$ both satisfy the diagram, by seifert-van kampen theorem, such map is unique. Therefore $\tilde{j}\circ \Phi=id_{\pi_1(X,x_0)}$. Consider $\Phi\circ \tilde{j}:\pi_1(U,x_0)* \pi_1(V,x_0)/N\to \pi_1(X,x_0)\to \pi_1(U,x_0)* \pi_1(V,x_0)/N$. _Motivation: $\Phi\circ\tilde j$ and $\mathrm{id}_H$ are both homomorphisms, so it suffices to check they agree on a generating set. The images of $\pi_1(U,x_0)$ and $\pi_1(V,x_0)$ under $q$ generate the quotient $H$, so we only need to check the identity on these two families._ $\forall g\in \pi_1(U,x_0)$, $\tilde{j}([g])=j(g)=j_1(g)$, and $\Phi(\tilde{j}([g]))=\Phi(j_1(g))=\phi_i(g)=[g]$. Similarly, $\forall h\in \pi_1(V,x_0)$, $\tilde{j}([h])=j(h)=j_2(h)$, and $\Phi(\tilde{j}([h]))=\Phi(j_2(h))=\phi_j(h)=[h]$. _Conclusion: $\Phi\circ\tilde j$ agrees with the identity on the generating classes from $\pi_1(U,x_0)$ and $\pi_1(V,x_0)$, so $\Phi\circ\tilde j=\mathrm{id}_H$. Together with $\tilde j\circ\Phi=\mathrm{id}_{\pi_1(X,x_0)}$, this proves $\tilde j$ is an isomorphism. $\square$_ </details> #### Proof of Seifert-Van Kampen Theorem It is really long, but it turns that it is somehow "similar" to the proof of $S^2$ case. Main idea is to cut the space into smaller pieces and decompose the loop into

f=f_1f_2f_3*\ldots*f_n

$$$$

\Phi(f)=\Phi_1(f_1)\Phi_2(f_2)\Phi_3(f_3)\ldots\Phi_n(f_n)

$$$$