§ CW Complexes and HEP

If XX is a CW complex and AA is a closed subcomplex, then it has the HEP. A closed subcomplex is a union of closed cells of XX such that XX is obtained by adding more cells to AA.

§ Lemma

If ee is a disk, then there is a continuous map from e×[0,1]e \times [0, 1] to e×[0,1](e×{0})\partial e \times [0, 1] \cup (e \times \{ 0 \}).

§ Lemma

If XX is obtained from AA by attaching one kk-cell, then (X,A)(X, A) has HEP. Given a homotopy ht:A×[0,1]Yh_t: A \times [0, 1] \rightarrow Y and a new homotopy F0:XYF_0: X \rightarrow Y such that F0A=htF_0|A = h_t, we want to complete FF such that FtA=htF_t|A = h_t. The only part I don't know where to define FF on is the new added ee portion. So I need to construct HH on e×[0,1]e \times [0, 1]. Use the previous map to get to e×[0,1](e×{0})e \times [0, 1] \cup (e \times \{0\}). This is in the domain of F0F_0 or hth_t, and thus we are done.

§ CW Complexes have HEP

Induction on lemma. base case is empty set.

§ Connected 1D CW Complex

Theorem: any connected 1D CW complex is homotopic to wedge of circles.
  • Find a contractible subcomplex AA of XX that passes through all 00 cells.
  • By HEP, XX/AX \simeq X/A. X/AX/A has only one zero-cell and other one cells. One cells are only attached to zero cells. Hence, is a wedge of circles.
  • The idea to find a contractible subcomplex is to put a partial order on the set of all contractible cell complexes by inclusion.
  • Pick a maximal element with respect to this partial order.
  • Claim: maximal element must contain all zero cells. Suppose not. Then I can add the new zero cell into the maximal element (why does it remain contractible? Fishy!)