Contractibility is product-closed

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This article gives the statement, and possibly proof, of a topological space property (i.e., contractible space) satisfying a topological space metaproperty (i.e., product-closed property of topological spaces)
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Property-theoretic statement

The property of topological spaces of being a contractible space, satisfies the metaproperty of topological spaces of being product-closed.

Statement with symbols

Let X_i, i \in I, be an indexed family of topological spaces. Then the product space, endowed with the product topology, is contractible.


Key idea (for two spaces)

Suppose F: X \times I \to X and G:Y \times I \to Y are contracting homotopies for X and Y. Then the map F \times G defined as:

(F \times G)(x,y,t) = (F(x,t),G(y,t))

is a contracting homotopy for X \times Y.

Thus X \times Y is contractible.

Generic proof (for an arbitrary family)

Given: An indexing set I, a collection \{ X_i \}_{i \in I} of contractible spaces. X is the product of the X_is, endowed with the product topology

To prove: X is a contractible space

Proof: Since each X_i is contractible, we can choose, for each X_i, a point p_i \in X_i, and a contracting homotopy F_i: X_i \times [0,1] \to X_i, with the property that:

F_i(a,0) = a \ \forall \ a \in X_i, F_i(a,1) = p_i \ \forall \ a \in X_i

Now consider the point p \in X whose i^{th} coordinate is p_i for each i \in I. We denote:

x = (x_i)_{i \in I}

to be a point whose i^{th} coordinate is x_i. Then, define a homotopy:

F: X \times [0,1] \to X

given by:

F(x,t) = (F_i(x_i,t))_{i \in I}

In other words, the homotopy acts as F_i in each coordinate. We observe that:

  • Since F_i(x_i,0) = x_i for each i, F(x,0) = x
  • Since F_i(x_i,1) = p_i for each i, F(x,1) = p
  • F is a continuous map: Fill this in later

Thus, F is a contracting homotopy on X, so X is contractible.