I'm reading the Bruzzo and Graña Otero's paper *Semistable and Numerically Effective Principal (Higgs) Bundles*; here: $X$ is a smooth, complex, projective variety; $G$ is a connected, complex, reductive, (affine) algebraic group; $E$ is a principal $G$-bundle on $X$.

A *Higgs field* $\varphi$ on $E$ is a global section of $Ad(E)\otimes_{\mathcal{O}_X}\Omega_X^1$ such that $[\varphi,\varphi]=0$. A *principal Higgs* $G$*-bundles on* $X$ is a pair $\mathfrak{E}=(E,\varphi)$.

Let $K$ be a closed subgroup of $G$ and let $\sigma:X\to E_{\displaystyle/K}$ be a reduction of the structure group of $E$ to $K$; that is, there exists a principal $K$-bundle $F_{\sigma}$ on $X$ and an injective $K$-equivariant bundle morphism $i_{\sigma}:F_{\sigma}\to E$. Let $\Pi_{\sigma}:Ad(E)\otimes_{\mathcal{O}_X}\Omega^1_X\to\left(Ad(E)_{\displaystyle/Ad(F_{\sigma})}\right)\otimes_{\mathcal{O}_X}\Omega^1_X$ be the induced projection; $\sigma$ is called a *Higgs reduction* of $\mathfrak{E}$ if and only if $\varphi\in\ker\Pi_{\sigma}$.

Let $E_{K}$ denotes the principal $K$-bundle $E\to E_{\displaystyle/K}$. Because:
\begin{gather*}
T_{E/K,X}\cong E_{\displaystyle/K}\left(Ad_K,\mathfrak{g}_{\displaystyle/\mathfrak{k}}\right),\\
\pi_K^{*}Ad(E)=\pi_K^{*}\left(E(Ad_G,\mathfrak{g})\right)=E_{\displaystyle/K}(Ad_K,\mathfrak{g}),
\end{gather*}
there exists a natural morphism $\eta:\pi_K^{*}Ad(E)\to T_{E/K,X}$ and $\varphi$ determines a global section $\eta(\varphi)\equiv(\eta\otimes Id)\left(\pi_K^{*}\varphi\right)$ of $T_{E/K,X}\otimes_{\mathcal{O}_{E/K}}\Omega^1_{E/K}$; the zero locus $\mathfrak{R}_K(\mathfrak{E})$ of $\eta(\varphi)$ is called *scheme of Higgs reduction of* $\mathfrak{E}$ (definition 3.5).

They state (page 6):

By construction, $\sigma:X\to E\left(G_{\displaystyle/K}\right)\cong E_{\displaystyle/K}$ is a Higgs reduction if and only if it takes value in the subscheme $\mathfrak{R}_K(\mathfrak{E})\subset E_{\displaystyle/K}$.

But I had check, via computation, that for any reduction $\sigma$ of $E$ one has $\mathfrak{R}_K(\mathfrak{E})$ is contained in the scheme-theoretic image of $\sigma$. Am I wrong?