Make the transfer sequence explicit in the case of a trivial covering \(\tilde{X} \to X\), where \(\tilde{X} = X \times S^0\).
Solution
Setup. The trivial covering \(p: X \times S^0 \to X\), \(p(x, \epsilon) = x\), is \(2\)-sheeted. The transfer \(\tau: C_n(X) \to C_n(X \times S^0)\) sends each singular simplex \(\sigma: \Delta^n \to X\) to the sum of its two lifts, and satisfies \(p_\# \tau = 2 \cdot \id\). The transfer sequence is the long exact sequence of the short exact sequence of chain complexes \[0 \to C_*(X) \xrightarrow{\tau} C_*(X \times S^0) \to C_*(X \times S^0)/\tau C_*(X) \to 0.\]
1. \(C_n(X \times S^0) \cong C_n(X) \oplus C_n(X)\).
\(X \times S^0 = X \sqcup X\) is the disjoint union of two copies of \(X\) (the sheets \(\epsilon = 0\) and \(\epsilon = 1\)), and singular chains split over disjoint unions.
2. Under 1, \(\tau\) is the diagonal map \(\Delta: C_n(X) \to C_n(X) \oplus C_n(X)\), \(\sigma \mapsto (\sigma, \sigma)\).
the two lifts of \(\sigma\) are \((\sigma, 0)\) and \((\sigma, 1)\), so \(\tau(\sigma) = (\sigma, 0) + (\sigma, 1) = (\sigma, \sigma)\) in the direct sum.
3. The quotient \(C_n(X \times S^0)/\tau C_n(X) \cong C_n(X)\).
3.1. \(\tau C_n(X) = \{(c, c) : c \in C_n(X)\}\).
2.
3.2. The map \((a, b) \mapsto a - b\) has kernel exactly \(\{(c, c)\}\) and is surjective.
\((a, b) \mapsto a - b\) vanishes iff \(a = b\); and \(a \mapsto (a, 0) \mapsto a\) shows surjectivity.
3.3. Hence \(C_n(X) \oplus C_n(X) / \{(c,c)\} \cong C_n(X)\) via the difference map.
first isomorphism theorem applied to 2.
4. The transfer sequence is therefore\[\cdots \to H_n(X) \xrightarrow{\Delta_*} H_n(X) \oplus H_n(X) \xrightarrow{\delta_*} H_n(X) \to H_{n-1}(X) \to \cdots\]where \(\Delta_*(x) = (x, x)\) and \(\delta_*(a, b) = a - b\).
apply the long exact sequence in homology to the short exact sequence of <1>1–<1>3.
5. The connecting homomorphisms \(H_n(X) \to H_{n-1}(X)\) are all zero.
\(\delta_*\) is surjective (since \((a, 0) \mapsto a\)), so the map out of \(H_n(X)\) in the sequence is zero.
6. Hence the transfer sequence splits into short exact sequences\[0 \to H_n(X) \xrightarrow{\Delta_*} H_n(X) \oplus H_n(X) \xrightarrow{\delta_*} H_n(X) \to 0\]for each \(n\).
4 and 5; the sequence is exact and the connecting maps vanish.
7. Q.E.D.
<1>4–<1>6 make the transfer sequence explicit: it is the split short exact sequence with diagonal inclusion and difference projection.