# Intersection pairing on $H_1$ of surfaces

Let $$X$$ be a surface (real, not complex). With coefficients in a ring $$R$$, there is an intersection pairing $$\langle -,- \rangle:H_1(X;\mathbb R) \otimes H_1(X;\mathbb R) \to R$$. Is this pairing perfect, or nondegenerate? I tend to get the terminology confused, but what I want to know is if the induced map $$H_1(X) \to \operatorname{Hom}(H_1(X),R)$$ is an isomorphism.

If $$X$$ is a closed surface, then my understanding is that the answer is yes. This is some guise of Poincare duality. But what if $$X$$ is noncompact, even infinite type (so that $$H_1$$ is not finite rank)? In particular, I'm curious about the case where $$X$$ is the infinite genus surface with one end. Here's a picture of that surface in the wild: Figure: Hernández, Jesús Hernández; Morales, Israel; Valdez, Ferrán, The Alexander method for infinite-type surfaces, Mich. Math. J. 68, No. 4, 743-753 (2019). ZBL1481.57038.

• Generally such a map won't be an isomorphism as the domain can be infinite-dimensional, so the two modules have ranks of different cardinalities. This map (the intersection pairing map) has finite support though, so you could restrict to that subspace of $Hom(H_1, R)$. Jun 8, 2022 at 1:27
• Ah thanks, I think I see now that the map is probably an embedding and that, via the universal coefficient isomorphism $\operatorname{Hom}(H_1(X),R) \cong H^1(X;R)$, the image is compactly supported cohomology. It's Poincare duality for noncompact manifolds. Jun 8, 2022 at 14:25

1. The intersection pairing is nondegenerate, i.e. the map $$i:H_1(X;R) \to \operatorname{Hom}(H_1(X;R),R)$$ is injective.
To see this, suppose that two homology classes $$\alpha$$ and $$\beta$$ have identical intersection numbers with all other homology classes. Since cycles are compact, we can take $$\alpha$$ and $$\beta$$ to live in some compact, $$H_1$$-injective subsurface with boundary $$X' \subset X$$. Then results about the intersection pairing on finite type surfaces show that $$\alpha$$ and $$\beta$$ are homologous in $$X'$$, and so they are homologous in $$X$$.
1. Under the isomorphism $$\operatorname{Hom}(H_1(X;R),R) \cong H^1(X;R)$$ coming from the universal coefficient theorem, the image of the map $$i$$ is $$H^1_c(X;R)$$.