TITLE: Why do subvarieties correspond to Hodge classes?
QUESTION [5 upvotes]: Let $X$ be a smooth complex projective variety and define
$$Hdg^k(X)=H^{2k}(X,\mathbb{Z})\cap H^{k,k}(X)$$
the group of integral $(k,k)$ cycles on $X$. Now it is a fact that we can 
associate to the complex subvariety of $X$ an element in $Hdg^{k}(X)$ but I don't quite get the details of the association. From what I have gathered so far I think we are working with the diagram
$$
\require{AMScd}
\begin{CD}
\mathcal{Z}_p(X) @>{i}>> H_{2n-2p}(X;\mathbb Z) @>{PD}>> H^{2p}(X;\mathbb Z)\\
@VVV   @. @VV{i}V \\
H^{2n-2p}_{dR}(X)^* @>{\cong}>> H^{2p}_{dR}(X) @>{de Rham}>> H^{2p}(X;\mathbb{C})
\end{CD}$$
Where along the top row we map
$$Z\mapsto [Z]\mapsto [Z]^{PD}$$
Where $^{PD}$ denotes the Poincare dual and $[Z]$ the pushforward of the fundamental class of $Z$ over the inclusion. Along the bottom row we map
$$Z\mapsto \left[\omega\mapsto \int_Zi^*\omega\right]\mapsto\left( \alpha \text{ s.t. } \int_{X}\beta\wedge \alpha=\int_Zi^*\beta\right)\mapsto \left(\alpha^{top} \text{ s.t. }\alpha^{top}(V)=\int_V\alpha\right)$$
(we assume that $Z$ is smooth and all that to make things easier). The last map is given by the inverse of the de Rham isomorphism.
Now the image of the bottom row can easily be shown to be a class of type $(p,p)$. The image of the top row is clearly the image of the integral cohomology class, by definition. 
Now if this diagram commutes we have a map $Z_p(X)\to Hdg^p(X)$. However, I don't see why this diagram should commute. So my question is: why does the above diagram commute?

I'll add the following link. At the bottom of page $148$ the authors states without proof (I renamed the objects to be consistent with the above):

The canonical morphism $H_{2n-2}(X,\mathbb{Z})\to H^{2n-2p}(X,\mathbb{C})^*$ carries the topological class $[Z]$ of an analytic subspace $Z$ of codimension $p$ in $X$ into the fundamental class $\left[\omega\mapsto \int_Zi^*\omega\right]$.
  Similarly, the morphism $H^{2p}(X,\mathbb{Z})\to H^{2p}(X,\mathbb{C})$ carries the topological class $[Z]^{PD}$ to $[\alpha^{top}]$

I am looking for a proof of this result. (this is equivalent with the diagram above commuting, but might give a bit more context)

REPLY [2 votes]: This is carried out in a lot of detail in Voisin – Hodge theory and complex algebraic geometry I, section 11.1.2. The particular result you're interested in seems to be Corollary 11.15.