answer:Let's denote the two boundary components as A and B. To find the intersection number of A and B using Poincaré duality, we first need to find the homology classes of A and B in H_1(X; Z), the first homology group of X with integer coefficients. Since X is a compact orientable surface of genus 2, it can be obtained from a 2-holed torus by removing two open disks. Let α, β, γ, and δ be the four simple closed curves on the 2-holed torus that generate H_1(X; Z). We can assume that α and β form the first hole, and γ and δ form the second hole. Now, let's consider the boundary components A and B. We can assume that A is the boundary of the first hole and B is the boundary of the second hole. Therefore, A can be represented as the homology class [α] + [β] and B can be represented as the homology class [γ] + [δ] in H_1(X; Z). To find the intersection number of A and B, we need to compute the cup product of their Poincaré duals in H^1(X; Z), the first cohomology group of X with integer coefficients. The Poincaré dual of A is PD(A) = α^* + β^*, and the Poincaré dual of B is PD(B) = γ^* + δ^*, where α^*, β^*, γ^*, and δ^* are the cohomology classes dual to α, β, γ, and δ, respectively. Now, we compute the cup product: (PD(A) ∪ PD(B))([X]) = (α^* + β^*) ∪ (γ^* + δ^*)([X]) = α^* ∪ γ^*([X]) + α^* ∪ δ^*([X]) + β^* ∪ γ^*([X]) + β^* ∪ δ^*([X]). Since α, β, γ, and δ are simple closed curves that do not intersect each other, their intersection numbers are all zero. Therefore, α^* ∪ γ^*([X]) = α^* ∪ δ^*([X]) = β^* ∪ γ^*([X]) = β^* ∪ δ^*([X]) = 0. Thus, the intersection number of A and B is: (PD(A) ∪ PD(B))([X]) = 0 + 0 + 0 + 0 = 0. So, the intersection number of the two boundary components A and B of the compact orientable surface of genus 2 is 0.

answer:To compute the intersection pairing, we will use the following facts: 1. The intersection pairing is bilinear, meaning it is linear in both arguments. 2. The intersection pairing is nondegenerate, meaning that for any nonzero element in H_1(A), there exists an element in H_1(B) such that their pairing is nonzero, and vice versa. Let's denote the generators of H_1(A) and H_1(B) as [a] and [b], respectively. Since H_1(A) is isomorphic to Z, we can write any element in H_1(A) as n[a] for some integer n. Similarly, any element in H_1(B) can be written as m[b] for some integer m, where m = 0 or 1 because H_1(B) is isomorphic to Z/2Z. Now, we want to compute the intersection pairing: H_1(A) x H_1(B) -> H_0(A ∩ B) We can do this by considering the pairing of the generators [a] and [b]: ⟨[a], [b]⟩ Since the pairing is nondegenerate, there must exist some nonzero integer k such that: ⟨[a], [b]⟩ = k Now, let's consider the pairing of [a] with 2[b]: ⟨[a], 2[b]⟩ = 2⟨[a], [b]⟩ = 2k However, since H_1(B) is isomorphic to Z/2Z, we know that 2[b] = 0 in H_1(B). Therefore, the pairing ⟨[a], 2[b]⟩ must be zero: ⟨[a], 2[b]⟩ = 0 This implies that 2k = 0. Since k is nonzero, we must be working in a group where 2 is equivalent to 0, which means that the intersection pairing takes values in Z/2Z. Thus, the intersection pairing can be described as: H_1(A) x H_1(B) -> H_0(A ∩ B) = Z/2Z with the property that ⟨[a], [b]⟩ = 1 in Z/2Z.

answer:To find the value of (-1)^{2(5-2)}, we first simplify the exponent: 2(5-2) = 2(3) = 6 Now, we can calculate the value: (-1)^6 = 1 So, the value of (-1)^{2(5-2)} in this case is 1.

answer:Let M be a compact 2-dimensional manifold with genus g geq 1. Let alpha and beta be loops with no common base-point that generate H_1(M; mathbb{Z}). We want to find the intersection number of the homology classes [alpha] and [beta]. Recall that Poincaré duality states that for a compact, oriented n-dimensional manifold M, there is an isomorphism PD: H_k(M; mathbb{Z}) to H^{n-k}(M; mathbb{Z}) for all k. In our case, n = 2, so we have an isomorphism PD: H_1(M; mathbb{Z}) to H^1(M; mathbb{Z}). Let PD([alpha]) = alpha^* and PD([beta]) = beta^*, where alpha^* and beta^* are the Poincaré dual cohomology classes in H^1(M; mathbb{Z}). To find the intersection number of [alpha] and [beta], we need to evaluate the cup product of their Poincaré duals on the fundamental class [M] in H_2(M; mathbb{Z}): langle alpha^* smile beta^*, [M] rangle = langle alpha^*, beta^* rangle. Now, since alpha and beta generate H_1(M; mathbb{Z}), they form a basis for the first homology group. In particular, they can be represented as simple closed curves on the surface M that intersect transversely in a single point. This means that their intersection number is pm 1, depending on the orientation of the intersection. Thus, the intersection number of the homology classes [alpha] and [beta] is langle alpha^*, beta^* rangle = pm 1.