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Is Being Hausdorff a Homotopy Invariant?

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Is Being Hausdorff a Homotopy Invariant? Topology Class Is being Hausdor a homotopy invariant? Yash Chandra Published on: Oct 08, 2020 License: Creative Commons Attribution 4.0 International License (CC-BY 4.0) Topology Class Is being Hausdor a homotopy invariant? Hi Everyone, In class, the concepts of homotopy and homotopy equivalence were defined. In that discussion, homotopy invariants – properties of a topological space that remain invariant under an homotopy equivalence – were introduced. Prof. Terilla left this as an open-ended question whether being Hausdorff is a homotopy invariant or not. We will try to address that question here. I’d like to thank Prof. Terilla for his valuable insights into this. I’d also like to acknowledge fellow Graduate Center students Ben, James, Moshe, and Ryan for informal discussions. The answer to the question is No, being Hausdorff is not a homotopy invariant! We will give a particular example to show that, and then try to give a generic recipe for getting more examples. Before doing anything technical, I think it is reasonable to guess that being Hausdorff is not a homotopy invariant, since homotopy is basically a topological deformation and one would intuitively expect that a non-Hausdorff space can be deformed into a Hausdorff one and vice-versa. Notice that we know that the one-point space is vacuously Hausdorff, so maybe we can find a homotopy equivalence between a non-Hausdorff space and the one-point space. We will try to exploit this idea here. A particular example: Consider the set X = {a, b} with the topology τ = {∅, {a}, X}. (X, τ ) is also known as the Sierpinski Space. Notice that (X, τ ) is not Hausdorff – a, b ∈ X cannot be separated by disjoint open sets because the only open set containing b is {a, b}. Next, we will show that this space is homotopy equivalent to the one-point space. Indeed, consider the map H : X × [0, 1] → X, where, H(a, 0) = a, H(b, 0) = b, H(a, t) = H(b, t) = a when t ∈ (0, 1]. Notice that H−1({a}) = ({a} × {0}) ∪ (X × (0, 1]) = ({a} × [0, 1]) ∪ (X × (0, 1]), is open in X × [0, 1] and hence H is continuous. From the definition of H, it is easy to see that H is a homotopy between idX and the constant map at a. Hence, (X, τ ) ≃ {a}. Now (X, τ ) is not Hausdorff while {a} is Hausdorff and hence we conclude that being Hausdorff is not a homotopy invariant. 2 Topology Class Is being Hausdor a homotopy invariant? A general example: The above particular example can be generalised in the following way: Consider a space for which there exists a point such that for all . (X, τ ) x ∈ X x ∈ U U ∈ τ , U = ∅ Notice that this implies that there are no pairs of non-empty disjoint open sets in X. Hence, (X, τ ) is not Hausdorff. Next we claim that X is homotopy equivalent to {x}. For that, consider the map H : X × [0, 1] → X, where, H(y, 0) = y for all y ∈ X; H(y, t) = x for all y ∈ X, for t ∈ (0, 1]. Then for an open set U in X, H−1(U ) = (U × {0}) ∪ (X × (0, 1]) = (U × [0, 1]) ∪ (X × (0, 1]), is open in and hence is continuous. By definition of , it is a homotopy between X × [0, 1] H H idX and the constant map at x. Hence, (X, τ ) ≃ {x}. As above, (X, τ ) is not Hausdorff while {x} is Hausdorff and hence this is an example of what we want. Okay, so we have shown that the above examples work. But how do we obtain a space like the one in our general example above? We will address that question now. Question: How to come up with such a space as above? Let (X, τX ) be any topological space. Let {a} denote the one-point space. Consider the set Y = X ∪ {a} and consider the collection τY = {∅} ∪ {U ∪ {a} : U ∈ τX }. It is easy to verify that τY defines a topology on Y , and by construction, a ∈ Y is in every non- empty open set in . Hence, is a space of the kind we were looking for. τY (Y , τY ) Conversely, it can be proved that every such space is obtained similarly in this way by adding one- point to a topological space. Indeed, let be a topological space satisfying the property of our (Y , τY ) general example, i.e. there exists a point such that for all . Consider x ∈ Y x ∈ U U ∈ τY , U = ∅ Z = Y ∖ {x} with the subspace topology induced from Y : τZ = {∅} ∪ {U ∖ {x} : U ∈ τY , U = ∅}. Now, let 3 Topology Class Is being Hausdor a homotopy invariant? W = {U ∖ {x} : U ∈ τY , U = ∅}, and G = {V ∪ {x} : V ∈ W }. Then, τY = ⟨G⟩, the topology generated by G. Therefore, we have now given a method of explicitly constructing a space like the one in our general example, starting from any topological space of our choice. Such a space will be non-Hausdorff and homotopy equivalent to the one-point space, thereby establishing that being Hausdorff is not a homotopy invariant. 4.
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