Pointed spaces and maps that preserve the basepoint
Objective: Why does specifying f(+1) determine a based map S⁰→X, while it does not determine an unbased map?
A pointed space is not merely a space with a decorative dot. It is a pair (X,x₀), and x₀ is part of the data. A based map f:(X,x₀)→(Y,y₀) must satisfy f(x₀)=y₀. This one equation determines which constant map is allowed, which homotopies count, and where later loop and smash constructions attach.
Write Map_*(X,Y) for based maps and [X,Y]_* for based homotopy classes. The star is a condition, not multiplication. A based homotopy H:X×I→Y keeps H(x₀,t)=y₀ for every t. An ordinary homotopy may move the image of x₀ and therefore need not prove equality in [X,Y]_*.
The zero object in pointed spaces is the one-point space *. There is exactly one based map *→X and exactly one based map X→*. This makes kernels, cones, wedges, and reduced constructions behave more like algebraic objects than their unbased counterparts, but it does not turn the category of pointed spaces into an abelian category.
The based zero-sphere S⁰ has two points {−1,+1} and basepoint −1. A based map S⁰→X is completely determined by the image of +1 because −1 is forced to x₀. Thus Map_*(S⁰,X) is naturally identified with the underlying set of points of X. This small model will later explain why S⁰ acts as the unit for smash product.
Whenever a formula uses a star, record three facts first: which object carries the basepoint, which map must preserve it, and whether the homotopy also preserves it at every time. Stable homotopy theory suppresses these facts typographically because they occur constantly; this course keeps them explicit until the type check becomes automatic.
For every pointed space (X,x₀), evaluation at the nonbasepoint +1 gives a natural bijection Map_*(S⁰,X)≅X. Under this bijection the constant based map corresponds to x₀.
Define ev₊(f)=f(+1). This is well typed because +1 is a point of S⁰ and f lands in X.
For x∈X, define f_x(−1)=x₀ and f_x(+1)=x. The domain is discrete, so f_x is continuous, and it is based by construction.
The composites satisfy ev₊(f_x)=x and f_{ev₊(f)}=f because a based map already has f(−1)=x₀. Naturality follows from h∘f_x=f_{h(x)} for every based h:X→Y.
Let X={x₀,a,b} be a discrete pointed space with basepoint x₀. List every based map S⁰→X and identify the constant based map.
- The basepoint condition forces f(−1)=x₀ in every case.
- The remaining point +1 may map independently to x₀, a, or b.
- Thus the maps are f_{x₀}, f_a, and f_b. Only f_{x₀} sends both domain points to x₀.
Result: There are exactly three based maps, corresponding to x₀,a,b under evaluation at +1; f_{x₀} is the constant based map.