Let be topological spaces, and let be two continuous maps. A homotopy between and is a continuous map
- , such that: .
(Note that the first argument of is usually denoted in subscript.) In the case of the existence of such an , the functions and are said to be homotopic to each other.
This means that is a "continuous transformation" of into or vice versa.
Definition (relative homotopy):
Let be a topological space, and let be a subset. If is another topological space, and if are continuous functions, then and are said to be homotopic relative (in short: ) if there exists a homotopy between and such that for all .
Proposition (relative homotopy is an equivalence relation):
If is a topological space, a subset and another topological space, then the relation
is an equivalence relation on .
Proof: First, note that by defining for , that is continuous, since for open, from which we gain reflexivity as also whenever . Symmetry follows upon considering, given a homotopy from to , the homotopy , which is continuous as the composition of with the continuous function ; indeed, the latter map is continuous since both of its components are; it is also a homotopy relative to , since for . Finally, note that transitivity holds, since if via the homotopy and via the homotopy , then via the homotopy
which is continuous since it is continuous on two closed sets that make up the whole space and relative to since both and are relative to .
Let be a topological space and let be a subset of . A retract of onto is a continuous function so that , the identity.
Definition (deformation retraction):
Let be a topological space and let be a subset of . A deformation retraction of onto is a continuous function so that for all (that is, ) and is a retract onto .
Note that the first argument is denoted in the lower index, so that instead of we write .
That is, a deformation retraction is a homotopy between the identity and a retract onto .
Definition (homotopy equivalence):
Let and be topological spaces. A homotopy equivalence between and is a pair of continuous functions and so that and .
Proposition (deformation retraction induces homotopy equivalence):
Let be a topological space, let be a subset, and let be a deformation retraction of onto . Then a homotopy equivalence is given by the pair and , being the inclusion.
Proof: We have , which is equal (and hence homotopic) to . Further, a homotopy between and is given by itself.
Let be a topological space. is called contractible iff there exists a point so that deformation retracts onto .
Proposition (contractible spaces are path-connected):
Let be a contractible space, that contracts to a point . Then any two points can be connected by a path that runs through .
Proof: For any point , if is the deformation retraction of onto , we get a path from to by the way of
which is continuous as the composition of continuous functions, since the map is continuous, since the first component is the identity and the second the constant function.
Let be a continuous function between topological spaces. is said to be nullhomotopic iff it is homotopic to a constant function, that is, to a function such that there exists so that for all .
Proposition (characterisation of contractibility):
Let be a topological space. The following are equivalent:
- is contractible
- The identity map on is nullhomotopic
- </math>X</math> is homotopy equivalent to a one-point space
- For every topological space , every continuous function is nullhomotopic
- For every topological space , any two continuous functions and are homotopic
Proof: Suppose first that is contractible, and let be the contraction. Then is simultaneously a homotopy between the identity and a constant map, where the latter has the value of the point onto which contracts. If the identity is homotopic to a constant map, and is the respective homotopy, then is also a deformation retraction and we conclude 3. since this deformation retraction induces a homotopy equivalence between a one-point space and . Let now be a topological space and a continuous function. Suppose further that and constitute a homotopy equivalence. Then in particular via some homotopy . But is a constant map. We then get via the homotopy , so that is nullhomotopic. Suppose now that and are two continuous functions. They will both be nullhomotopic, and if we can show that is path-connected and is the constant function to which is homotopic and for any , then they will both be homotopic to , since is homotopic to any constant map via , where is a curve from to the point of the other constant map . But is path-connected, since upon choosing , the two-point space with discrete topology, we get that the (continuous) function sending and to two points is nullhomotopic, and the homotopy yields a path between and by the way of
- , where and .
Finally, if any two continuous functions from any space to are homotopic, we may as well choose and one map to be any constant map and the other one the identity; the homotopy between the two then yields the desired deformation retraction.
Definition (homotopy extension property):
Let be a topological space and a subset. The pair is said to have the homotopy extension property if and only if for every topological space and every continuous function such that is homotopic to some other map via a homotopy , there exists a homotopy so that and .
Proposition (characterisation of the homotopy extension property via retraction):
Let be a topological space and a subset. has the homotopy extension property if and only if there exists a retraction from onto .
Proof: Suppose first that does have the homotopy extension property. Then pick and define by . Then is homotopic to the function via , and then by the homotopy extension property we get a homotopy of to some other function . By the form of , restricts to the identity on , and further it maps to , so that is a retraction. Suppose now that is a retract of via the function , where we define to ease notation. Let (where is some topological space) be a function and a homotopy so that . We define
and claim that it is continuous. Note that is the composition of (which is continuous) with the map
so that it suffices to prove that is continuous. To this end, it will be sufficient to prove that whenever is a set such that and are open, then itself is open; indeed, if this is the case, then whenever is an open subset of , we'll have that
is open. So suppose that has said property. We show that is open by fixing a and finding an open neighbourhood of in that is contained within . If , then we may just choose a suitable product neighbourhood since cylinders are a basis of the product topology and by the definition of the subspace topology. If and , we may just choose the set . The only remaining case is the case and . If that is the case, suppose first that , and consider the point . We write the retraction as . Since and for all , we get that , since denoting the neighbourhood filter of by , we get that converges to by continuity of (which follows from the characterisation of continuity of functions to a space with initial topology), but is contained in each set , being a neighbourhood of , since , by definition of the product topology, always contains a point of the form , where , and we conclude since is Hausdorff.
For each , we define to be the maximal open subset of so that ; a maximal such set exists since sets having this property are closed under union, and we take the union over all the sets having this property. Then define
is then an open subset of . Now , since otherwise, for all we have . However, then for arbitrary and , we get that , since if , then by continuity of , we find and open with , so that , which implies in particular that . However, for , we simply have since is a retraction, and hence we get and . Hence, since , and thus implies . However, note that still ; indeed, by what was proven above, we have , and then, since is a retraction whence , we must have . We conclude that for all . Now we observe that if is the projection from to , then , since whenever so that , we find an open and an such that , and we then write , where is open, and obtain . By continuity of , we then obtain , since the preimage of the complement of will be closed in . But , so that in fact , a contradiction.
Hence, choose so that ; then the neighbourhood will do as an open neighbourhood of contained within , using the definitions of the subspace and product topologies.