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is path connected

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Here we refine the argument of the previous section to prove is path connected. There are two main difficulties that arise. One is that the path connected analogue of Lemma 4.2, although still true (at least when M is Hausdorff), is much harder to prove. The second is that a decreasing intersection of compact path connected sets need not be path connected, so we can no longer restrict our attention to the zeros within .
The lifting lemma below will be used as a substitute for Lemma 4.2. Its proof is based on proofs obtained independently by David desJardins and Emanuel Knill.

Lemma 5.1

  (Lifting lemma): Let M be a Hausdorff space and let be the projection map. Let be a continuous map. Then there is a continuous map such that .

SubLemma 5.1

  Let consists of n copies of a single point. Let be an arbitrary function that is a lift of f. Then g is automatically continuous at all .

[Proof]

SubLemma 5.2

  Let , be closed subintervals of such that is a single point . If is a continuous lift of f on then there is a continuous lift g of f on such that .

[Proof]

SubLemma 5.3

  The conclusions of Sublemma 5.2 hold even if and intersect in more than a point.

[Proof]

SubLemma 5.4

  If I is a closed subinterval of and every has a neighborhood on which f has a lift, then f has a lift on I.

[Proof]

SubLemma 5.5

  The same holds if I is any subinterval of .

[Proof]

[Proof]

Theorem 5.1

  is path connected.

[Proof]



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