NoteNextra-origin/content/Math4202/Math4202_L16.md

Math4202 Topology II (Lecture 16)

Algebraic Topology

Fundamental group of the circle

Recall from previous lecture, we have unique lift for covering map.

Lemma for unique lifting for covering map

Let p: E\to B be a covering map, and e_0\in E and p(e_0)=b_0. Any path f:I\to B beginning at b_0, has a unique lifting to a path starting at e_0.

Back to the circle example, it means that there exists a unique correspondence between a loop starting at (1,0) in S^1 and a path in \mathbb{R} starting at 0, ending in \mathbb{Z}.

Proof

\forall t\in I, by the covering map, partition into slices property, there exist some open neighborhood U_{f(t)} of f(t) such that p^{-1}(U_{f(t)})\subseteq E is a neighborhood of e_0. And p^{-1}(U_{f(t)}) is a disjoint union of \{V_{f(t),\alpha}\}

Since f:I\to B is continuous, then V_t of t\in I is open in I and we can find some small open neighborhood f^{-1}(V_t)\subseteq U_{f(t)}.

Note that \{V_t\} is an open cover of [0,1]. As [0,1] is compact, \{V_t\} has a finite subcover, \{V_{t_i}\}_{i=1}^k.

Then we can use \{V_{t_i}\}_{i=1}^k to partition I into 0<s_1<s_2<\cdots<s_m=1, such that [s_i,s_{i+1}] is contained in one of V_{t_i}.

We can do so by consider the [0,t_1)\subseteq V_{t_1}, therefore [t_1,1]\subseteq \bigcup_{i=2}^k V_{t_i}. (These open sets should intersect with [t_1,1] non trivially, no useless choice shall be made.)

Since \bigcup_{i=2}^k V_{t_i} is open, there is z_1<t_1 and (z_1,1]\subseteq \bigcup_{i=2}^k V_{t_i}.

Then we can choose z_1<s_1<t_1, since s_1\in \bigcup_{i=2}^k V_{t_i}, there exists some V_{t_2} such that s_1\in V_{t_2}.

Note that we can find [0,t_2)\subseteq V_{t_1}\cup V_{t_2}, and t_2>t_1.

Continue this process, we can find our partition 0<s_1<s_2<\cdots<s_m=1, such that [s_i,s_{i+1}] is contained in one of V_{t_i}.

In conclusion, we find the above partition of I such that

f([s_i,s_{i+1}])\subset f(V_{t_i})\subseteq U_i

Define \tilde{f}:I\to E, on [0,s_1], f([0,s_1])\subseteq U_0, p^{-1}(U_0)=\bigcup_\alpha W_{0,\alpha}, \exists \alpha_0 such that e_0\in W_{0,\alpha_0}.

\tilde{f}=(p^{-1}|_{W_{s_i,\alpha_i}})\circ f

Repeat the same construction for each [s_i,s_{i+1}].

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The uniqueness for lifting map is guaranteed by the following:

• For e_0, W_{0,\alpha_0}\to U_0 is a homeomorphism, suppose \tilde{f} and \tilde{f}' are liftings of f, then they must agree on \tilde{f}([0,s_1])=\tilde{f}'([0,s_1])=p^{-1}(U_0)\circ f.
• Repeat the same construction for each [s_i,s_{i+1}].

Lemma for unique lifting homotopy for covering map

Let p: E\to B be a covering map, and e_0\in E and p(e_0)=b_0. Let F:I\times I\to B be continuous with F(0,0)=b_0. There is a unique lifting of F to a continuous map \tilde{F}:T\times I\to E, such that \tilde{F}(0,0)=e_0.

Further more, if F is a path homotopy, then \tilde{F} is a path homotopy.

Discuss on Firday.