[proofplan]
We realize cohomology classes by closed complex-valued differential forms and use the smooth pullback of forms. The pullback of forms commutes with the [exterior derivative](/theorems/1525) and with the wedge product, so it descends to a degree-preserving graded ring homomorphism on de Rham cohomology. Holomorphicity gives the Hodge-type compatibility by [citetheorem:8075]. Finally, the cycle class of a closed complex submanifold has pure Hodge type by [citetheorem:8074], and the type-compatibility just proved preserves that type after pullback.
[/proofplan]
[step:Descend pullback of differential forms to a graded cohomology homomorphism]
For each integer $k\geq 0$, let $A^k(M;\mathbb C)$ denote the complex [vector space](/page/Vector%20Space) of smooth complex-valued $k$-forms on a smooth manifold $M$. Since $f:X\to Y$ is holomorphic, it is smooth, so for every $k\geq 0$ the usual pullback of differential forms defines a $\mathbb C$-[linear map](/page/Linear%20Map)
\begin{align*}
f^\sharp_k:A^k(Y;\mathbb C)\to A^k(X;\mathbb C).
\end{align*}
This pullback commutes with the exterior derivative:
\begin{align*}
d(f^\sharp_k\alpha)=f^\sharp_{k+1}(d\alpha)
\end{align*}
for every $\alpha\in A^k(Y;\mathbb C)$.
Let $\alpha\in A^k(Y;\mathbb C)$ be closed, meaning $d\alpha=0$. Then
\begin{align*}
d(f^\sharp_k\alpha)=f^\sharp_{k+1}(d\alpha)=0,
\end{align*}
so $f^\sharp_k\alpha$ is closed. If $\alpha=d\beta$ for some $\beta\in A^{k-1}(Y;\mathbb C)$, then
\begin{align*}
f^\sharp_k\alpha=f^\sharp_k(d\beta)=d(f^\sharp_{k-1}\beta),
\end{align*}
so exact forms pull back to exact forms. Therefore the assignment
\begin{align*}
[\alpha]\mapsto [f^\sharp_k\alpha]
\end{align*}
is well-defined on de Rham cohomology and gives a $\mathbb C$-linear map
\begin{align*}
f^*:H^k(Y;\mathbb C)\to H^k(X;\mathbb C).
\end{align*}
[guided]
The first point is that the cohomological pullback is not introduced abstractly; it is induced by pullback of differential forms. For a smooth manifold $M$, let $A^k(M;\mathbb C)$ be the complex vector space of smooth complex-valued $k$-forms on $M$. Since a holomorphic map is smooth, the map $f:X\to Y$ has a usual smooth pullback on forms, so for each integer $k\geq 0$ we have a $\mathbb C$-linear map
\begin{align*}
f^\sharp_k:A^k(Y;\mathbb C)\to A^k(X;\mathbb C).
\end{align*}
To pass from forms to cohomology classes, we must check two things: closed forms pull back to closed forms, and exact forms pull back to exact forms. The standard naturality identity for the exterior derivative says that
\begin{align*}
d(f^\sharp_k\alpha)=f^\sharp_{k+1}(d\alpha)
\end{align*}
for every $\alpha\in A^k(Y;\mathbb C)$. If $\alpha$ is closed, then $d\alpha=0$, and hence
\begin{align*}
d(f^\sharp_k\alpha)=f^\sharp_{k+1}(0)=0.
\end{align*}
Thus $f^\sharp_k\alpha$ is closed.
Now suppose $\alpha$ is exact. This means that there exists a form $\beta\in A^{k-1}(Y;\mathbb C)$ such that $\alpha=d\beta$. Applying the same naturality identity gives
\begin{align*}
f^\sharp_k\alpha=f^\sharp_k(d\beta)=d(f^\sharp_{k-1}\beta).
\end{align*}
Thus $f^\sharp_k\alpha$ is exact. Consequently, replacing a closed representative $\alpha$ by a cohomologous representative changes $f^\sharp_k\alpha$ only by an exact form, so the formula
\begin{align*}
[\alpha]\mapsto [f^\sharp_k\alpha]
\end{align*}
is well-defined. This is precisely the ordinary cohomological pullback
\begin{align*}
f^*:H^k(Y;\mathbb C)\to H^k(X;\mathbb C).
\end{align*}
[/guided]
[/step]
[step:Use wedge-product naturality to prove the graded ring property]
Let $a\in H^k(Y;\mathbb C)$ and $b\in H^\ell(Y;\mathbb C)$, and choose closed representatives $\alpha\in A^k(Y;\mathbb C)$ and $\beta\in A^\ell(Y;\mathbb C)$ such that $a=[\alpha]$ and $b=[\beta]$. The cup product in de Rham cohomology is represented by the wedge product:
\begin{align*}
a\smile b=[\alpha\wedge\beta].
\end{align*}
The pullback of differential forms commutes with wedge products:
\begin{align*}
f^\sharp_{k+\ell}(\alpha\wedge\beta)=f^\sharp_k\alpha\wedge f^\sharp_\ell\beta.
\end{align*}
Therefore
\begin{align*}
f^*(a\smile b)=[f^\sharp_{k+\ell}(\alpha\wedge\beta)]=[f^\sharp_k\alpha\wedge f^\sharp_\ell\beta]=f^*a\smile f^*b.
\end{align*}
Since $f^\sharp_k$ sends $k$-forms to $k$-forms, the induced map sends $H^k(Y;\mathbb C)$ to $H^k(X;\mathbb C)$. Hence
\begin{align*}
f^*:H^\bullet(Y;\mathbb C)\to H^\bullet(X;\mathbb C)
\end{align*}
is a graded $\mathbb C$-algebra homomorphism.
[/step]
[step:Apply holomorphic Hodge-type preservation to cohomology classes]
Because $X$ and $Y$ are compact Kähler manifolds and $f:X\to Y$ is holomorphic, [citetheorem:8075] applies to $f$. Its conclusion gives, for every pair of integers $p,q\geq 0$,
\begin{align*}
f^*\bigl(H^{p,q}(Y)\bigr)\subset H^{p,q}(X).
\end{align*}
This is exactly the asserted compatibility of $f^*$ with the [Hodge decomposition](/theorems/2745).
[/step]
[step:Pull back the cycle class of a closed complex submanifold]
Let $Z\subset Y$ be a closed complex submanifold of complex codimension $r$, and let
\begin{align*}
[Z]\in H^{2r}(Y;\mathbb C)
\end{align*}
be its Poincaré dual cohomology class. Since $Y$ is a compact Kähler manifold and $Z$ is a closed complex submanifold of codimension $r$, [citetheorem:8074] applies and gives
\begin{align*}
[Z]\in H^{r,r}(Y)\subset H^{2r}(Y;\mathbb C).
\end{align*}
Applying the Hodge-type preservation from the previous step with $p=r$ and $q=r$ yields
\begin{align*}
f^*[Z]\in f^*\bigl(H^{r,r}(Y)\bigr)\subset H^{r,r}(X).
\end{align*}
Since $f^*$ preserves cohomological degree, $f^*[Z]$ lies in $H^{2r}(X;\mathbb C)$. Thus the ordinary cohomological pullback of $[Z]$ is a class of pure type $(r,r)$, hence a Hodge class of the same cohomological degree in the stated sense.
[/step]
