# Topology lecture notes by Ward T.

By Ward T.

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1. Lifting maps We next turn to the following problem: if p : Z → X is a covering map, and f : Y → X is a map, when can we expect there to be a lift f of f ; that is a map f : Y → Z such that pf = f . 1. 5. Painting sides of a surface It will be convenient to adopt the following convention: a commutative diagram of the form f Y −−−→ Z  p f Y −−−→ X should be thought of as a triangle. 5. Let p : (Z, z0 ) → (X, x0 ) be a based covering map, and f : Y → X a map from a connected space. Then, if there is a lift f of (a map making the diagram above commute), it is unique.

To start you off, notice that in rule [2] (cancellation of . . aa−1 . . we lose 1 edge and 1 vertex, preserving χ. The other rules are similar. Under barycentric subdivision, let K be the combinatorial surface obtained from K. Then α0 (K ) = α0 (K) + α1 (K) + α2 (K) (one new vertex for each edge and each triangle in K), α1 (K ) = 2α1 (K) + 6α2 (K) (six new edges in each triangle of k and each edge of k split into two), and α2 (K ) = 6α2 (K). It follows that χ(K ) = α0 (K ) − α1 (K ) + α2 (K ) = α0 (K) + α1 (K) + α2 (K) − 2α1 (K) − 6α2 (K) + 6α2 (K) = α0 (K) − α1 (K) + α2 (K) = χ(K).

BXddb−1 · · · −→ . . bXe−1 be−1 . . , and then by [3], . . bXe−1 be−1 · · · −→ . . bbeX −1 e−1 . . That is, we have reduced the original pair separated by d1 d1 . . dn dn dd to a similar pair separated by −1 −1 −1 d−1 n dn . . d1 d1 . 3. 20. Pairs of the kind . . b . . b−1 . . After n such steps, the symbol has the form . . bddb−1 · · · −→ . . be−1 be−1 . . (by the same cut and paste with X = ∅), and then . . be−1 be−1 · · · −→ . . f f . . by [1], with f = be−1 . Notice that the existence of such a pair .