Action of an isomorphism in cohomology as the intersection with the class of the graph Let $X$ and $Y$ be two complex manifolds of dimension 2 and let $\varphi:X\rightarrow Y$ be an isomorphism. 
I have read that the action of $\varphi^*:H^2(Y,\mathbb{Z})\rightarrow H^2(X,\mathbb{Z})$ can be seen as $\omega\mapsto {p_1}_*([\Gamma]\cdot p_2^*(\omega))$ for all $\omega\in H^2(Y,\mathbb{Z})$. Here $p_1$ and $p_2$ are the two projections from $X\times Y$, $\Gamma$ is the graph of $\varphi$ and i interpret ${p_1}_*([\Gamma]\cdot p_2^*(\omega))$ as $\mathscr{P}({p_1}_*(\mathscr{P}^{-1}(\mathscr{P}(\Gamma)\wedge p_2^*(\omega))))$, where $\mathscr{P}$ is the Poincarè duality.
I think i can see why this is working, but do you know a formal proof for that or do you know where i can find it?
Thank you
 A: I believe this is a special case of a more general fact; I am not sure of all the signs off the top of my head, but here  is the idea. 
If $M$ and $N$ are orientable $d$-manifolds, the Künneth theorem gives
$$H^d(M \times N; \mathbb{Q}) \cong \bigoplus_k H^k(M; \mathbb{Q}) \otimes H^{d-k}(N; \mathbb{Q}).$$
To the second factor we first apply Poincare duality $H^{d-k}(N; \mathbb{Q}) \cong H_k(N; \mathbb{Q})$ and then the usual duality $H_k(N;\mathbb{Q})\cong H^k(N;\mathbb{Q})^\ast$ to rewrite this as
$$H^d(M \times N; \mathbb{Q}) \cong \bigoplus_k \text{Hom}\big( H^k(N; \mathbb{Q}), H^k(M; \mathbb{Q}) \big).$$
For any continuous map $f\colon M \to N$, the graph $\Gamma_f$ of the map $f$ is a $d$-dimensional submanifold of $M \times N$, so we can consider its class
$$[\Gamma_f] \in H^d(M \times N; \mathbb{Q}).$$
Claim: under the above isomorphism, we have
$$[\Gamma_f] = \bigoplus_k \big[\ f_{(k)}^*\colon H^k(N; \mathbb{Q}) \to H^k(M; \mathbb{Q}) \big]$$
where $f_{(k)}^*$ denotes the map induced on $k$-th cohomology by $f$.
There might be some signs missing here, but other than that I believe this is correct. (I don't know a reference for this, so if anyone does have a reference, I'd be grateful.) In particular, it doesn't seem that you need to assume that $\phi$ is an isomorphism, only continuous, or that $M$ and $N$ are surfaces.
As an aside, if you apply this twice with $M=N$, once to $\text{id}\colon M\to M$ and once to $f\colon M\to M$, you should be close to proving the Lefschetz fixed point theorem.

Edit: as David Speyer points out, this aside requires knowing that intersection of submanifolds is Poincare dual to the cup product. A reference for this can be found in the textbook "Differential forms in algebraic topology" by Bott and Tu, in 6.31 (on page 69): "under Poincare duality the transversal intersection of closed oriented submanifolds corresponds to the wedge product of forms". (From one perspective, this is the most important thing to know about cup product!)
Also, Mark Grant has kindly provided a reference for the fact I claimed above: this is proved in his answer to this Math Overflow question from 2010, by reducing the claim to the case of the diagonal embedding (i.e. when $f=\text{id}$).
A: This is proved for the Chow ring in Fulton's "Intersection Theory", Section 16.1. 
In topology it can be proved using the following facts (which it may be easier to find references for individually). Let $X$ and $Y$ be smooth manifolds.


*

*A proper embedding $e: M\hookrightarrow Y$ of codimension $k$ with oriented (stable) normal bundle represents a cohomology class $[e]\in H^k(Y;\mathbb{Z})$. Call such a class representable.

*Every element of $H^2(Y;\mathbb{Z})$ is representable.

*Given a map $f: X\to Y$, the induced map $f^\ast: H^\ast(Y;\mathbb{Z})\to H^\ast(X;\mathbb{Z})$ is given on representable classes by pulling back a representing embedding by a smooth transverse map homotopic to $f$.

*If $g:X\to Y$ is a proper map with oriented stable normal bundle, the pushforward $g_\ast: H^\ast(X;\mathbb{Z})\to H^\ast(Y;\mathbb{Z})$ is given on representable classes by composing a representative with $g$.

*If $[f]\in H^k(Y;\mathbb{Z})$ and $[e]\in H^\ell(Y;\mathbb{Z})$ are represented by submanifolds, then their product $[f]\cdot [e]\in H^{k+\ell}(X; \mathbb{Z})$ equals $f_\ast f^\ast[e]$.


The formula you give follows on noting that $\Gamma_\varphi:X\to X\times Y$ satisfies $p_1\circ\Gamma_\varphi = \operatorname{id}_X$ and $p_2\circ\Gamma_\varphi = \varphi$.
