# General Topology/Homotopy

**Definition (homotopy)**:

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 .

**Definition (retract)**:

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.

**Definition (contractible)**:

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.

**Definition (nullhomotopic)**:

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.