# Close maps are homotopic

## 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