Close maps are homotopic

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Statement

Suppose X is a topological space and Y is a compact subset in Euclidean space, such that there exists an open subset U \supset Y such that Y is a strong deformation retract of U. Then, there exists \epsilon > 0 such that if two maps f_1,f_2:X \to Y are \epsilon-close, in the sense:

d(f_1(x), f_2(x)) < \epsilon

(where d denotes the Euclidean distance) then f_1 and f_2 are homotopic.

Alternative interpretations

A concrete interpretation of this is as follows. Suppose we view Y as a compact metric space with the metric induced as a subset of \R^n. Then we can give C(X,Y) the topology of uniform convergence. There is a natural map:

C(X,Y) \to [X,Y]

where [X,Y] denotes the space of homotopy classes of continuous maps from X to Y. The above result says that the above map is continuous if we give [X,Y] the discrete topology. This interpretation follows because for every function in a homotopy class, the \epsilon-neighbourhood of that function is also in the same homotopy class.

The advantage of this interpretation is that for Y a compact metric space, the topology of uniform convergence coincides with the compact-open topology, which can be defined without reference to the explicit metric. Thus, we can state the result more abstractly as:

If X is a topological space and Y is a compact metrizable space, give C(X,Y) the compact-open topology and [X,Y] the discrete topology. Then the mapping:

C(X,Y) \to [X,Y]

is continuous.

Converse

A converse to this statement exists, but under different hypotheses; we need to assume that the space X is compact and Y just needs to be a metric space. Fill this in later