[proofplan] Choose the given Kähler form $\omega$ as a closed positive real $(1,1)$-form representative of the class $[\omega]$. The pullback of differential forms commutes with the [exterior derivative](/theorems/1525), so $f^*\omega$ is closed and represents the cohomological pullback $f^*[\omega]$. Holomorphicity gives $df_x\circ J_X=J_Y\circ df_x$, which proves that the pullback form has type $(1,1)$ and lets us test semipositivity by evaluating $\omega$ on the vector $df_x(v)$ and its $J_Y$-rotation. The possible vanishing of $df_x(v)$ is exactly why the pulled-back form is semipositive rather than necessarily positive. [/proofplan]

[step:Choose a Kähler representative of the class] Since $\omega$ is a Kähler form on $Y$, it is a smooth real closed positive $(1,1)$-form. Its de Rham cohomology class is the given class \begin{align*} [\omega]\in H^{1,1}(Y)\cap H^2(Y,\mathbb R). \end{align*} We will prove that the smooth pullback form $f^*\omega$ has the asserted properties on $X$. [/step]

[step:Show that the pulled-back form is closed and represents the pulled-back cohomology class] The exterior derivative is natural with respect to smooth pullback, so \begin{align*} d(f^*\omega)=f^*(d\omega). \end{align*} Since $\omega$ is closed, $d\omega=0$, hence \begin{align*} d(f^*\omega)=0. \end{align*} Therefore $f^*\omega$ is a closed real two-form on $X$, and by the definition of the induced map on de Rham cohomology it represents \begin{align*} [f^*\omega]=f^*[\omega]\in H^2(X,\mathbb R). \end{align*} The compactness and Kähler hypotheses on $X$ ensure that the compact Kähler [Hodge decomposition](/theorems/2745) applies to $X$ [citetheorem:8066]. Thus, once we verify that the representative $f^*\omega$ has type $(1,1)$, this real de Rham class is an element of $H^{1,1}(X)\cap H^2(X,\mathbb R)$. [/step]

[step:Use holomorphicity to preserve type and reality]For each $x\in X$, let \begin{align*} df_x:T_xX\to T_{f(x)}Y \end{align*} denote the differential of $f$ at $x$. Since $f$ is holomorphic, its differential is complex-linear: \begin{align*} df_x(J_Xv)=J_Ydf_x(v) \end{align*} for every $x\in X$ and every $v\in T_xX$. Let $x\in X$ and let $u,v\in T_xX$. Because $\omega$ has type $(1,1)$ as a real two-form on $Y$, it satisfies \begin{align*} \omega_{f(x)}(J_Ya,J_Yb)=\omega_{f(x)}(a,b) \end{align*} for all $a,b\in T_{f(x)}Y$. Using the definition of pullback and the complex-linearity of $df_x$, we obtain \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(df_x(J_Xu),df_x(J_Xv)). \end{align*} Thus \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(J_Ydf_x(u),J_Ydf_x(v)). \end{align*} Since $\omega$ has type $(1,1)$, \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(df_x(u),df_x(v))=(f^*\omega)_x(u,v). \end{align*} Hence $f^*\omega$ has type $(1,1)$ on $X$. Since $\omega$ is real and pullback commutes with complex conjugation on differential forms, $f^*\omega$ is real. Therefore \begin{align*} f^*\omega\in A^{1,1}(X) \end{align*} as a real form. Combining this type computation with the closedness proved above and the compact Kähler [Hodge decomposition](/theorems/3941) on $X$ [citetheorem:8066], its class lies in $H^{1,1}(X)\cap H^2(X,\mathbb R)$.[/step]

[guided]The point of this step is to convert holomorphicity of $f$ into a statement about the type decomposition of forms. For each $x\in X$, the differential of $f$ is the real-[linear map](/page/Linear%20Map) \begin{align*} df_x:T_xX\to T_{f(x)}Y. \end{align*} Because $f$ is holomorphic, this real-linear map is compatible with the two complex structures: \begin{align*} df_x(J_Xv)=J_Ydf_x(v) \end{align*} for every $v\in T_xX$. Now fix $x\in X$ and tangent vectors $u,v\in T_xX$. The pullback form is defined by inserting the pushed-forward tangent vectors into $\omega$: \begin{align*} (f^*\omega)_x(u,v)=\omega_{f(x)}(df_x(u),df_x(v)). \end{align*} To check that $f^*\omega$ has type $(1,1)$ as a real two-form, we test invariance under applying $J_X$ to both arguments. Using the definition of pullback gives \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(df_x(J_Xu),df_x(J_Xv)). \end{align*} Holomorphicity replaces each occurrence of $df_x(J_X\cdot)$ by $J_Ydf_x(\cdot)$, so \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(J_Ydf_x(u),J_Ydf_x(v)). \end{align*} Since $\omega$ is a real $(1,1)$-form on $Y$, it satisfies \begin{align*} \omega_{f(x)}(J_Ya,J_Yb)=\omega_{f(x)}(a,b) \end{align*} for all $a,b\in T_{f(x)}Y$. Applying this with $a=df_x(u)$ and $b=df_x(v)$ gives \begin{align*} (f^*\omega)_x(J_Xu,J_Xv)=\omega_{f(x)}(df_x(u),df_x(v)). \end{align*} By the definition of pullback, the right-hand side is \begin{align*} (f^*\omega)_x(u,v). \end{align*} Therefore $f^*\omega$ has type $(1,1)$. Reality is also preserved under pullback: since $\omega$ takes real values on real tangent vectors, the form $f^*\omega$ takes real values on real tangent vectors. Hence $f^*\omega$ is a real $(1,1)$-form on $X$. Together with the closedness already proved, the compact Kähler Hodge decomposition on $X$ [citetheorem:8066] identifies the de Rham class of this closed real $(1,1)$-form as an element of \begin{align*} H^{1,1}(X)\cap H^2(X,\mathbb R). \end{align*}[/guided]

[step:Verify semipositivity by evaluating on complex lines] Let $x\in X$ and $v\in T_xX$. Using the definition of pullback and holomorphicity of $f$, we compute \begin{align*} (f^*\omega)_x(v,J_Xv)=\omega_{f(x)}(df_x(v),df_x(J_Xv)). \end{align*} Since $df_x(J_Xv)=J_Ydf_x(v)$, this becomes \begin{align*} (f^*\omega)_x(v,J_Xv)=\omega_{f(x)}(df_x(v),J_Ydf_x(v)). \end{align*} The form $\omega$ is positive because it is Kähler, so for every $w\in T_{f(x)}Y$, \begin{align*} \omega_{f(x)}(w,J_Yw)\geq 0. \end{align*} Applying this with $w=df_x(v)$ gives \begin{align*} (f^*\omega)_x(v,J_Xv)\geq 0. \end{align*} Thus $f^*\omega$ is semipositive. Combining this with closedness, reality, and type $(1,1)$ proves that $f^*[\omega]$ is represented by the closed semipositive real $(1,1)$-form $f^*\omega$. [/step]