Consider the first singular homology and cohomology groups of a modular curve, $H^1(X,\mathbb{Z})$ and $H_1(X,\mathbb{Z})$. The Hecke algebra acts on both of them and they are dual to each other under the Poincaré pairing. Hence we get an isomorphism $H_1(X,\mathbb{Z})\cong Hom(H^1(X,\mathbb{Z}),\mathbb{Z})$.
Do you know if it is equivariant for the action of the Hecke algebra? If so can you give me the reference?
Thank you.
The easy answer: Yes, there is a natural definition of Hecke operators on $H_1$ that makes this work. This is explained in Stein's book, as Qiaochu has already pointed out, and in lots of other places too.
But there is a more subtle phenomenon going on here. The isomorphism $H_1 \cong Hom(H^1, \mathbf{Z})$ doesn't really come from Poincare duality (it's immediate from the universal coefficient theorem, once you know that the homology is torsion-free). What Poincare duality gives you is the isomorphism $H_1 \cong H^1$; and there is no way to define the Hecke actions that will make both isomorphisms $H_1 \cong H^1$ and $H_1 \cong (H^1)^*$ be Hecke-equivariant at the same time.
If you regard Poincare duality as a pairing $H^1 \times H^1 \to \mathbf{Z}$ (rather than in terms of homology) then the statement is that the Hecke operators are not selfadjoint with respect to this pairing. One can write $T_p$ as $\alpha_* \circ \beta^*$ where $\alpha$ and $\beta$ are two maps from some higher level modular curve $X'$ to $X$; and from the compatibility of pushforwards and cup-products, the transpose of $T_p$ is $T_p' = \beta_* \circ \alpha^*$, which isn't generally the same.
One can check that if the level is something like $\Gamma_1(N)$ and $p \nmid N$, then $T_p' = \langle p \rangle^{-1} T_p$ (or maybe $\langle p \rangle T_p$, I'd have to check). This is a familiar calculation with $\mathbf{C}$ coefficients, where the Poincare duality pairing is basically the Petersson scalar product, but it works with $\mathbf{Z}$ coefficients as well.