n维球面到拓扑空间映射的等价性证明问询（Hatcher拓扑拓展）

Great question—this generalization works flawlessly for every non-negative integer (n), and the proof mirrors the core reasoning from the (n=1) case you referenced in Hatcher's Algebraic Topology section 1.1.5. Let's walk through the equivalence step by step.

First, let's fix notation: (S^n) denotes the (n)-dimensional sphere, (D^{n+1}) the ((n+1))-dimensional closed disk, with the critical relation (\partial D^{n+1} = S^n) (the boundary of the disk is exactly the sphere).

Proving (b) ⇒ (a): Extendable maps are nullhomotopic #

Suppose (f: S^n \to X) extends to a continuous map (\tilde{f}: D^{n+1} \to X) (so (\tilde{f}\big|_{S^n} = f)). We can construct a homotopy from (f) to a constant map directly:

• Define (H: S^n \times [0,1] \to X) by (H(x,t) = \tilde{f}(tx)).
• When (t=1), (H(x,1) = \tilde{f}(x) = f(x)) (matches our original map).
• When (t=0), (H(x,0) = \tilde{f}(0)) for all (x \in S^n)—this is a constant map (all points map to the image of the disk's center).
• Continuity of (H) follows immediately from continuity of (\tilde{f}) (scaling (x) by (t) is continuous, and composition of continuous maps is continuous).

So (f) is homotopic to a constant map, satisfying condition (a).

Proving (a) ⇒ (b): Nullhomotopic maps extend to the disk #

Now suppose (f: S^n \to X) is nullhomotopic—meaning there's a continuous homotopy (H: S^n \times [0,1] \to X) where (H(x,1) = f(x)) and (H(x,0) = x_0) for some fixed (x_0 \in X) (the constant map). We need to build an extension (\tilde{f}: D^{n+1} \to X).

Key observation: The disk (D^{n+1}) is the quotient space of (S^n \times [0,1]) where we collapse the entire set (S^n \times {0}) to a single point (the origin of the disk). The quotient map here is (q: S^n \times [0,1] \to D^{n+1}) given by (q(x,t) = tx).

Since (H) maps every point in (S^n \times {0}) to the same (x_0), it descends to a continuous map (\tilde{f}: D^{n+1} \to X) (by the universal property of quotient spaces: if a map is constant on each equivalence class of a quotient map, it induces a continuous map on the quotient). For any (x \in S^n), (\tilde{f}(x) = \tilde{f}(q(x,1)) = H(x,1) = f(x)), so (\tilde{f}) is indeed an extension of (f) to (D^{n+1}), satisfying condition (b).

A quick side note: This equivalence is deeply tied to the definition of the (n)-th homotopy group (\pi_n(X)). Condition (a) is equivalent to (\pi_n(X)) being the trivial group (for path-connected (X); for non-path-connected (X), this holds for each path component individually).

内容的提问来源于stack exchange，提问作者amir bahadory