Hatcher, *Algebraic Topology*, Chapter 0

**16. Show that is contractible.**

*Proof.* Dr. Hatcher himself gives the proof of this in Example 1B.3 on Page 88, a proof the ideas of which are replicated here.

The goal is to show that is contractible, i.e. that the identity map is homotopic to a constant map. It will be shown that where is the map for all . First, note that is the collection of all points for which : This fact will be used to build the desired homotopy, the construction of which will take place in two parts.

Let be the homotopy

for all . This is a straight-line homotopy between the point and the point from which we define the restriction for all . Clearly, then, is a homotopy between points and . Next, let be given by

.

As with , is a straight-line homotopy, this time between the points and . As above, defining the restriction for all gives a chain of homotopies

.

In particular, then, the composition is the desired homotopy between and the constant map , where

.

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I’m interested why you need to apply the straight line homotopy f first, instead of just using g.

Best,

Lee

Given $f_t (X)(1-t)(x_1,x_2…..) + t(1,0,0,0…..)$, wouldnt $F=f_t(X) / |f_t(X)|$ be enough?