Suppose $M^m$ and $N^n$ are smooth, connected, closed, oriented manifolds, with $m\ge n$.If we have a smooth map $f:M\rightarrow N$, we can define a graph as the submanifold $\Gamma=\{(x, f(x))\mid x\in M\}\subset M\times N$.
Is there an explicit description of the Poincare dual $\eta_\Gamma\in H^n(M\times N)$ of $\Gamma$?
Here, I'm thinking about a Poincare dual in the sense of Bott-Tu: it is the equivalence class $[\eta_\Gamma]\in H^n(M\times N)$ of a closed $n$-form $\eta_\Gamma$ with the property that, if $[\omega]\in H^m(M\times N)$ is any $m$-form, there is an equality $$ \int_\Gamma \omega_{|\Gamma}=\int_{M\times N}\omega\wedge\eta_\Gamma$$
My gut instinct was that $\eta_\Gamma=\pi_2^\ast\alpha$, where $\alpha\in H^n(N)$ is nonzero, and $\pi_2:M\times N\rightarrow N$ is projection. But this seems wrong, since it doesn't even mention $f$.
However, by taking $\omega=\pi_1^\ast\beta$ above (where $\beta\in H^m(M)$ is nonzero), it would seem that $\eta_\Gamma=\pi_2^\ast\alpha+\cdots$.
Is there an explicit representation of $\eta_\Gamma$ in terms of $\alpha$, $\beta$, and the various maps mentioned earlier?
After @TedShifrin gave me the hint in the comments, I followed the exposition in Bott-Tu, and came up with the following:
$$ \eta_\Gamma=\sum_{\deg(\alpha_i)+\deg(\beta_j)=n} c_{ij}\pi_1^\ast\alpha_i\wedge\pi_2^\ast\beta_j$$
where $\alpha_i$ are a basis for $H^\ast(M)$ and $\beta_j$ are a basis for $H^\ast(N)$. If $\overline{\alpha}_i$ denotes a Poincare-dual basis to $\alpha_i$, and $\overline{\beta}_j$ denotes a Poincare-dual basis to $\beta_j$, then we can write the $c_{ij}$ as
$$ c_{ij} = (-1)^{\deg(\overline{\beta}_j)}\int_M\overline{\alpha}_i\wedge f^\ast\overline{\beta}_j$$
This seems to agree with the case that $\Gamma=\Delta$ is the diagonal, and $f$ is the identity map from $M$ to $N=M$.
I'll also mention this agrees with the case I'm more familiar with, where $N=S^2$. Then if $\beta\in H^2(S^2)$ is the volume form, we should have $$ \eta_\Gamma=\pi_1^\ast f^\ast\beta + \pi_2^\ast\beta $$ and this agrees with the formula above.