Intersection pairing on $H_1$ of surfaces

Question

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.

Answer 1

Thanks to Ryan Budney in the comments for jogging my memory. Here's what I think can be said:

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 $X$.

2. 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 $H^1_c(X;R)$.

I'm pretty sure this is just Poincare duality for noncompact manifolds.