Let \(A\) be a subspace of \(\mathbb{R}^n\); let \(h: (A, a_0) \to (Y, y_0)\). Show that if \(h\) is extendable to a continuous map of \(\mathbb{R}^n\) into \(Y\), then \(h_*\) is the trivial homomorphism (the homomorphism that maps everything to the identity element).
Solution
Let \(H: \mathbb{R}^n \to Y\) be a continuous extension of \(h\).
1. Factorization through the ambient space \(\mathbb{R}^n\):Proof:
1.1. Let \(i: (A, a_0) \hookrightarrow (\mathbb{R}^n, a_0)\) denote the canonical inclusion map.
1.2. Since \(H: (\mathbb{R}^n, a_0) \to (Y, y_0)\) is a continuous extension of \(h\), the composition satisfies:\[h = H \circ i: (A, a_0) \to (Y, y_0).\]
1.3. By functoriality of the fundamental group operator \(\pi_1\):\[h_* = (H \circ i)_* = H_* \circ i_*,\]where \(i_*: \pi_1(A, a_0) \to \pi_1(\mathbb{R}^n, a_0)\) and \(H_*: \pi_1(\mathbb{R}^n, a_0) \to \pi_1(Y, y_0)\).
2. Simple connectivity of \(\mathbb{R}^n\):Proof:
2.1. The Euclidean space \(\mathbb{R}^n\) is convex, and the straight-line homotopy \(K(x, t) = (1 - t)x + t a_0\) exhibits \(\mathbb{R}^n\) as contractible to \(a_0\).
2.2. In particular, \(\mathbb{R}^n\) is simply connected, so its fundamental group is trivial:\[\pi_1(\mathbb{R}^n, a_0) = \{1\}.\]
3. Triviality of the homomorphism \(h_*\):Proof:
3.1. Let \([\gamma] \in \pi_1(A, a_0)\) be an arbitrary element.
3.2. Since \(\pi_1(\mathbb{R}^n, a_0) = \{1\}\), the image \(i_*([\gamma])\) must be the identity element \(1 \in \pi_1(\mathbb{R}^n, a_0)\).
3.3. Because group homomorphisms preserve the identity element:\[h_*([\gamma]) = (H_* \circ i_*)([\gamma]) = H_*(1) = 1 \in \pi_1(Y, y_0).\]
3.4. Thus \(h_*([\gamma]) = 1\) for all \([\gamma] \in \pi_1(A, a_0)\).
4. Q.E.D.Step 3 shows that \(h_*\) is trivial.