# Intersection pairing on non-compact surface

Let $$S$$ be a smooth oriented connected $$2$$-manifold. We have an algebraic intersection pairing $$\omega\colon H_1(S) \times H_1(S) \rightarrow \mathbb{Z}$$. If $$S$$ is compact, then this is nondegenerate in the sense that it identifies $$H_1(S)$$ with its dual $$\text{Hom}(H_1(S),\mathbb{Z})$$. If $$S$$ is not compact, then this rarely holds as defined since for instance if $$H_1(S) \cong \mathbb{Z}^{\infty}$$, then $$H_1(S)$$ is countable but $$\text{Hom}(H_1(S),\mathbb{Z})$$ is uncountable.

This suggests re-defining "non-degenerate" as follows. Define $$\text{Hom}_c(H_1(S),\mathbb{Z})$$ to be the subspace of $$\text{Hom}(H_1(S),\mathbb{Z})$$ consisting of $$f\colon H_1(S) \rightarrow \mathbb{Z}$$ such that there exists a compact subsurface $$T$$ of $$S$$ such that $$f$$ vanishes on all cycles supported outside of $$T$$. Note that for all $$x_0 \in H_1(S)$$ the map $$\omega(x_0,-)$$ lies in $$\text{Hom}_c(H_1(S),\mathbb{Z})$$, and we say that $$\omega$$ is nondegenerate if it identifies $$H_1(S)$$ with $$\text{Hom}_c(H_1(S),\mathbb{Z})$$.

This sometimes holds and sometimes does not hold. Here are five examples:

1. If $$S$$ is a once-punctured genus $$g$$ surface, then the intersection pairing is nondegenerate.

2. If $$S$$ is a multiply-punctured genus $$g$$ surface, then the intersection pairing is degenerate.

3. If $$S$$ is a sphere minus a Cantor set, then the intersection pairing is identically $$0$$, and in particular is degenerate.

4. If $$S$$ is an infinite genus surface with one end (think, the usual picture of a genus $$g$$ surface but with infinitely many holes going off to the right), then the intersection pairing is nondegenerate.

5. If $$S$$ is an infinite genus surface with two ends (think, the usual picture of a genus $$g$$ surface but with infinitely many holes going off to both the left and the right), then the intersection pairing is degenerate.

What is the general case? The above examples suggest that the right condition is "has at most one end", but I'm not sure how to prove something like that.

Recall the non-compact Poincar'e duality: $$H^*(M)\cong \bar H_{m-*}(M), H^*_c(M)\cong H_{m-*}(M).$$ Here $$M$$ is a $$\mathbb{Z}$$-orientable topological manifold without boundary, $$\bar H$$ is the Borel-Moore homology, $$H^*_c$$ is the compactly supported cohomology, $$m=\dim M$$, and all coefficients are integral. The pairing is obtained by using the second of these isomorphisms and the cup product assuming $$M$$ path connected.
If $$M$$ is not compact, the pairing is rarely non-degenerate. Case 1 above follows from the fact that we have a ring map $$H^\ast_c(S)\to H^\ast(\bar S)$$ where $$\bar S$$ is the one point compactification, and it so happens that $$\bar S$$ is a topological manifold and the map is an isomorphism in degree 1. Case 4 is not really an example: in this case $$H_1$$ is free Abelian whereas $$Hom(H_1,\mathbb{Z})$$ is not.
Upd: regarding the new version of the question. Let us generalize the definition given above by setting $$Hom_c(H_p(M),\mathbb{Z})=$$ the image of the map $$H_{m-p}(M)\to H^p_c(M)\to H^p(M)\to Hom(H_p(M),\mathbb{Z}).$$ Here all coefficients are again integral. The last two arrows make sense for an arbitrary space, and the first one is the Poincar'e isomorphism.
Assuming $$M$$ path connected, all groups in the above sequence can be paired with $$H_p(M)$$ in a compatible way. For the first group we get the pairing $$H_{m-p}(M)\times H_p(M)\to\mathbb{Z}$$ we're after. For the last group we get just the evaluation pairing. The image of $$H_{m-p}(M)\to Hom (H_p(M),\mathbb{Z})$$ induced by the pairing $$H_{m-p}(M)\times H_p(M)\to\mathbb{Z}$$ is the same as the image of the above composition, i.e. $$Hom_c(H_p(M),\mathbb{Z})$$.
So the question we want to answer can be rephrased as: when is the image equal $$H_{m-p}(M)$$, i.e. when is the composition injective? And the answer is somewhat tautological: the composition is injective iff the preimage of $$Ext(H_{p-1}(M),\mathbb{Z})\subset H^p(M)$$ in $$H^p_c(M)$$ is zero. If $$m=2, p=1$$ the Ext vanishes, and the condition is equivalent to $$H^1_c(M)\to H^1(M)$$ being injective.