[proofplan] We prove functoriality of the [Hodge decomposition](/theorems/2745) through the filtered de Rham complex of differential forms. First, a local coordinate computation shows that a holomorphic map pulls back forms of type $(p,q)$ to forms of type $(p,q)$. Since pullback commutes with the [exterior derivative](/theorems/1525), it sends each filtered subcomplex of forms with holomorphic degree at least $r$ into the corresponding filtered subcomplex on the source. The compact Kähler [Hodge decomposition](/theorems/3941) identifies these filtrations with the usual Hodge filtrations on cohomology, and compatibility with complex conjugation then forces each pure summand $H^{p,q}$ to map into the corresponding summand. [/proofplan]

[step:Show that holomorphic pullback preserves the type of smooth forms]Let $m=\dim_{\mathbb C}Y$ and $n=\dim_{\mathbb C}X$. For a complex manifold $M$, let $A^k(M;\mathbb C)$ denote the complex [vector space](/page/Vector%20Space) of smooth complex-valued differential $k$-forms on $M$, and let \begin{align*} A^\bullet(M;\mathbb C):=\bigoplus_{k\geq 0}A^k(M;\mathbb C) \end{align*} denote the graded complex vector space of all smooth complex-valued differential forms on $M$. Let $\Lambda^{p,q}T^*M\to M$ denote the complex vector bundle of covariant differential forms of type $(p,q)$, and define \begin{align*} A^{p,q}(M):=\Gamma(M,\Lambda^{p,q}T^*M). \end{align*} Thus $A^{p,q}(Y)$ and $A^{p,q}(X)$ are the complex vector spaces of smooth complex-valued differential forms of type $(p,q)$ on $Y$ and $X$, respectively. We prove that the smooth pullback \begin{align*} f^*:A^\bullet(Y;\mathbb C)\to A^\bullet(X;\mathbb C) \end{align*} satisfies \begin{align*} f^*A^{p,q}(Y)\subset A^{p,q}(X) \end{align*} for all $p,q\geq 0$. Fix a point $x\in X$, and put $y=f(x)$. Choose a holomorphic coordinate chart $(U,\varphi)$ on $X$ with $x\in U$ and holomorphic coordinates $(z_1,\dots,z_n)$, and choose a holomorphic coordinate chart $(V,\psi)$ on $Y$ with $y\in V$ and holomorphic coordinates $(w_1,\dots,w_m)$, after shrinking $U$ so that $f(U)\subset V$. In these coordinates, write the coordinate expression of $f|_U$ as \begin{align*} F:=\psi\circ f\circ \varphi^{-1}:\varphi(U)\to \psi(V)\subset \mathbb C^m. \end{align*} Let $F_j:\varphi(U)\to \mathbb C$ denote the $j$-th component function of $F$ for $1\leq j\leq m$. Since $f$ is holomorphic, each $F_j$ is holomorphic, so its Wirtinger derivatives satisfy \begin{align*} \frac{\partial F_j}{\partial \bar z_a}=0 \end{align*} for every $1\leq a\leq n$. Hence, on $U$, \begin{align*} f^*(dw_j)=d(F_j\circ\varphi)=\sum_{a=1}^n \frac{\partial F_j}{\partial z_a}\,dz_a. \end{align*} Thus $f^*(dw_j)$ is a form of type $(1,0)$. Taking complex conjugates gives \begin{align*} f^*(d\bar w_j)=\sum_{a=1}^n \overline{\frac{\partial F_j}{\partial z_a}}\,d\bar z_a, \end{align*} so $f^*(d\bar w_j)$ is a form of type $(0,1)$. Every local section of $\Lambda^{p,q}T^*Y$ is a finite smooth linear combination of wedge products with exactly $p$ factors among the $dw_j$ and exactly $q$ factors among the $d\bar w_j$. Since pullback is multiplicative with respect to wedge products, the pullback of such a wedge product has exactly $p$ factors of type $(1,0)$ and $q$ factors of type $(0,1)$ on $X$. Therefore $f^*$ sends $A^{p,q}(Y)$ into $A^{p,q}(X)$.[/step]

[guided]We want to prove type preservation at the level where it is genuinely visible: in holomorphic coordinates. Let $m=\dim_{\mathbb C}Y$ and $n=\dim_{\mathbb C}X$, and let $A^{p,q}(Y)$ and $A^{p,q}(X)$ denote the spaces of smooth complex-valued forms of type $(p,q)$ on $Y$ and $X$. Fix $x\in X$ and set $y=f(x)$. Choose a holomorphic chart $(U,\varphi)$ on $X$ around $x$ with holomorphic coordinates $(z_1,\dots,z_n)$, and choose a holomorphic chart $(V,\psi)$ on $Y$ around $y$ with holomorphic coordinates $(w_1,\dots,w_m)$. Since $f$ is continuous, we may shrink $U$ to an open neighbourhood of $x$ satisfying $f(U)\subset V$. The coordinate representative of $f$ is the holomorphic map \begin{align*} F:=\psi\circ f\circ \varphi^{-1}:\varphi(U)\to \psi(V)\subset \mathbb C^m. \end{align*} For each $1\leq j\leq m$, let \begin{align*} F_j:\varphi(U)\to \mathbb C \end{align*} be the $j$-th component of $F$. Holomorphicity means precisely that the anti-holomorphic Wirtinger derivatives vanish: \begin{align*} \frac{\partial F_j}{\partial \bar z_a}=0 \end{align*} for all $1\leq a\leq n$. Now compute the pullback of the basic $(1,0)$-forms on $Y$. Since $w_j\circ f=F_j\circ\varphi$ on $U$, the chain rule gives \begin{align*} f^*(dw_j)=d(F_j\circ\varphi)=\sum_{a=1}^n \frac{\partial F_j}{\partial z_a}\,dz_a+\sum_{a=1}^n \frac{\partial F_j}{\partial \bar z_a}\,d\bar z_a. \end{align*} The second sum vanishes by holomorphicity, so \begin{align*} f^*(dw_j)=\sum_{a=1}^n \frac{\partial F_j}{\partial z_a}\,dz_a. \end{align*} This is a linear combination of $(1,0)$-forms on $X$, hence has type $(1,0)$. Taking complex conjugates gives \begin{align*} f^*(d\bar w_j)=\sum_{a=1}^n \overline{\frac{\partial F_j}{\partial z_a}}\,d\bar z_a, \end{align*} which is a linear combination of $(0,1)$-forms on $X$, hence has type $(0,1)$. A general local $(p,q)$-form on $Y$ is a finite smooth linear combination of wedge products with exactly $p$ differentials of the form $dw_j$ and exactly $q$ differentials of the form $d\bar w_j$. Pullback respects smooth coefficients and wedge products. Therefore every such local wedge product pulls back to a wedge product with exactly $p$ factors of type $(1,0)$ and exactly $q$ factors of type $(0,1)$. Since type is checked locally, this proves \begin{align*} f^*A^{p,q}(Y)\subset A^{p,q}(X) \end{align*} for all $p,q\geq 0$.[/guided]

[step:Use naturality of $d$ to make pullback commute with $\partial$ and $\bar\partial$] The smooth pullback of differential forms commutes with the exterior derivative: \begin{align*} d(f^*\alpha)=f^*(d\alpha) \end{align*} for every smooth complex-valued differential form $\alpha$ on $Y$. Let $\alpha\in A^{p,q}(Y)$. By the [type decomposition of the exterior derivative](/theorems/7004) from [citetheorem:7004], \begin{align*} d\alpha=\partial\alpha+\bar\partial\alpha, \end{align*} where $\partial\alpha\in A^{p+1,q}(Y)$ and $\bar\partial\alpha\in A^{p,q+1}(Y)$. Using the previous step, $f^*(\partial\alpha)$ has type $(p+1,q)$ and $f^*(\bar\partial\alpha)$ has type $(p,q+1)$. Hence \begin{align*} d(f^*\alpha)=f^*(\partial\alpha)+f^*(\bar\partial\alpha) \end{align*} is the type decomposition of $d(f^*\alpha)$. By uniqueness of the direct-sum decomposition into types, we obtain \begin{align*} \partial(f^*\alpha)=f^*(\partial\alpha) \end{align*} and \begin{align*} \bar\partial(f^*\alpha)=f^*(\bar\partial\alpha). \end{align*} [/step]

[step:Descend the type-preserving pullback to Dolbeault cohomology] For each $p,q\geq 0$, define the Dolbeault cohomology group \begin{align*} H_{\bar\partial}^{p,q}(Y):=\frac{\ker(\bar\partial:A^{p,q}(Y)\to A^{p,q+1}(Y))}{\operatorname{im}(\bar\partial:A^{p,q-1}(Y)\to A^{p,q}(Y))} \end{align*} with the convention $A^{p,-1}(Y)=0$, and define $H_{\bar\partial}^{p,q}(X)$ analogously. The identities $\bar\partial^2=0$ needed for these quotients are part of the [Dolbeault operator relations](/theorems/8046) in [citetheorem:8046]. Let $\alpha\in A^{p,q}(Y)$ satisfy $\bar\partial\alpha=0$. By commutation with $\bar\partial$, \begin{align*} \bar\partial(f^*\alpha)=f^*(\bar\partial\alpha)=0, \end{align*} so $f^*\alpha$ is $\bar\partial$-closed. If $\alpha$ is changed by a $\bar\partial$-exact form, say \begin{align*} \alpha'=\alpha+\bar\partial\beta \end{align*} for some $\beta\in A^{p,q-1}(Y)$, then \begin{align*} f^*\alpha'=f^*\alpha+f^*(\bar\partial\beta)=f^*\alpha+\bar\partial(f^*\beta). \end{align*} Thus $f^*\alpha'$ and $f^*\alpha$ define the same class in $H_{\bar\partial}^{p,q}(X)$. Therefore pullback defines a complex-[linear map](/page/Linear%20Map) \begin{align*} f_{\bar\partial}^{p,q}:H_{\bar\partial}^{p,q}(Y)\to H_{\bar\partial}^{p,q}(X) \end{align*} by \begin{align*} f_{\bar\partial}^{p,q}([\alpha]_{\bar\partial})=[f^*\alpha]_{\bar\partial}. \end{align*} [/step]

[step:Show that de Rham pullback preserves the Hodge filtration]Fix an integer $k\geq 0$. For every integer $r$, define the filtered space of smooth complex-valued $k$-forms on a complex manifold $M$ by \begin{align*} F^rA^k(M;\mathbb C):=\bigoplus_{a+b=k,\,a\geq r}A^{a,b}(M). \end{align*} The type decomposition of the exterior derivative from [citetheorem:7004] gives $d=\partial+\bar\partial$, where $\partial$ raises the first type index by $1$ and $\bar\partial$ preserves the first type index. Therefore \begin{align*} d\bigl(F^rA^k(M;\mathbb C)\bigr)\subset F^rA^{k+1}(M;\mathbb C), \end{align*} so $F^rA^\bullet(M;\mathbb C)$ is a subcomplex of the complex de Rham complex $A^\bullet(M;\mathbb C)$. The first step proves that smooth pullback sends $A^{a,b}(Y)$ into $A^{a,b}(X)$ for every $a,b\geq 0$. Hence it sends $F^rA^k(Y;\mathbb C)$ into $F^rA^k(X;\mathbb C)$ for every $k$ and $r$. Since smooth pullback also commutes with $d$, it is a morphism of filtered complexes \begin{align*} f^*:(A^\bullet(Y;\mathbb C),F^\bullet)\to (A^\bullet(X;\mathbb C),F^\bullet). \end{align*} Passing to cohomology, the ordinary de Rham pullback satisfies \begin{align*} f^*F^rH^k(Y;\mathbb C)\subset F^rH^k(X;\mathbb C) \end{align*} for every integer $r$, where for compact Kähler manifolds the Hodge filtration is identified by [Hodge Decomposition for Compact Kähler Manifolds](citetheorem:TEMP-24) as \begin{align*} F^rH^k(M;\mathbb C)=\bigoplus_{a+b=k,\,a\geq r}H^{a,b}(M). \end{align*}[/step]

[guided]The subtle point is that an arbitrary Dolbeault class cannot simply be treated as a de Rham class represented by the same form: if $\alpha\in A^{p,q}(Y)$ satisfies $\bar\partial\alpha=0$, it may still have $\partial\alpha\neq 0$, so it need not be $d$-closed. We avoid that mistake by working with the Hodge filtration on the de Rham complex itself. For a complex manifold $M$ and an integer $r$, define \begin{align*} F^rA^k(M;\mathbb C):=\bigoplus_{a+b=k,\,a\geq r}A^{a,b}(M). \end{align*} This is the space of smooth $k$-forms whose components have holomorphic degree at least $r$. The type decomposition theorem [citetheorem:7004] says that $d=\partial+\bar\partial$, with $\partial:A^{a,b}(M)\to A^{a+1,b}(M)$ and $\bar\partial:A^{a,b}(M)\to A^{a,b+1}(M)$. Thus $d$ cannot decrease the first type index. Consequently \begin{align*} d\bigl(F^rA^k(M;\mathbb C)\bigr)\subset F^rA^{k+1}(M;\mathbb C), \end{align*} and $F^rA^\bullet(M;\mathbb C)$ is a subcomplex of the complex de Rham complex. The local coordinate computation already proved the essential functorial input: because $f$ is holomorphic, $f^*$ sends each $A^{a,b}(Y)$ into $A^{a,b}(X)$. Therefore it sends the direct sum of all components with $a\geq r$ into the corresponding direct sum on $X$: \begin{align*} f^*F^rA^k(Y;\mathbb C)\subset F^rA^k(X;\mathbb C). \end{align*} Since pullback of smooth forms commutes with the exterior derivative, this is a morphism of filtered complexes \begin{align*} f^*:(A^\bullet(Y;\mathbb C),F^\bullet)\to (A^\bullet(X;\mathbb C),F^\bullet). \end{align*} Passing to de Rham cohomology gives \begin{align*} f^*F^rH^k(Y;\mathbb C)\subset F^rH^k(X;\mathbb C). \end{align*} For compact Kähler manifolds, [Hodge Decomposition for Compact Kähler Manifolds](citetheorem:TEMP-24) identifies this filtration with \begin{align*} F^rH^k(M;\mathbb C)=\bigoplus_{a+b=k,\,a\geq r}H^{a,b}(M). \end{align*} This proves filtration preservation for the ordinary de Rham pullback, which is the compatibility needed for the Hodge decomposition.[/guided]

[step:Recover preservation of each pure Hodge summand] Let $p,q\geq 0$ and put $k=p+q$. By [Hodge Decomposition for Compact Kähler Manifolds](citetheorem:TEMP-24), the compact Kähler Hodge structures on $H^k(Y;\mathbb C)$ and $H^k(X;\mathbb C)$ are opposed, so \begin{align*} H^{p,q}(M)=F^pH^k(M;\mathbb C)\cap \overline{F^qH^k(M;\mathbb C)} \end{align*} for $M=Y$ and $M=X$. The rational pullback is defined over $\mathbb Q$, so its complexification commutes with complex conjugation on cohomology. The previous step gives preservation of $F^p$, and after conjugating the same filtration preservation gives preservation of $\overline{F^q}$. Therefore, for every class $\gamma\in H^{p,q}(Y)$, \begin{align*} f^*\gamma\in F^pH^k(X;\mathbb C)\cap \overline{F^qH^k(X;\mathbb C)}=H^{p,q}(X). \end{align*} Thus \begin{align*} f^*H^{p,q}(Y)\subset H^{p,q}(X). \end{align*} [/step]

[step:Conclude that the rational pullback is a morphism of pure Hodge structures] Fix an integer $k\geq 0$. The continuous map underlying $f$ induces a rational cohomology pullback \begin{align*} f^*:H^k(Y;\mathbb Q)\to H^k(X;\mathbb Q). \end{align*} After extension of scalars from $\mathbb Q$ to $\mathbb C$, this is the complex de Rham pullback \begin{align*} f^*:H^k(Y;\mathbb C)\to H^k(X;\mathbb C). \end{align*} The previous step proves that this complexified map satisfies \begin{align*} f^*H^{p,q}(Y)\subset H^{p,q}(X) \end{align*} for every pair $p,q\geq 0$ with $p+q=k$. Equivalently, if $F^rH^k(Y;\mathbb C)$ and $F^rH^k(X;\mathbb C)$ denote the Hodge filtrations defined by \begin{align*} F^rH^k(Y;\mathbb C):=\bigoplus_{\substack{p+q=k, p\geq r}}H^{p,q}(Y) \end{align*} and \begin{align*} F^rH^k(X;\mathbb C):=\bigoplus_{\substack{p+q=k, p\geq r}}H^{p,q}(X), \end{align*} then \begin{align*} f^*F^rH^k(Y;\mathbb C)\subset F^rH^k(X;\mathbb C) \end{align*} for every integer $r$. This is precisely the condition that the rational pullback be a morphism of pure Hodge structures of weight $k$. This proves both the well-definedness of the maps on $H^{p,q}$ and the asserted Hodge-structure compatibility. [/step]