Consider an operator $\eta$ on the (tensor) product of Hilbert spaces $\mathcal{H}_1\otimes \mathcal{H}_2$ such that its range is of the form $V_1\otimes V_2$. Does this imply that $\eta$ must be a product operator, i.e., of the form $\eta = A\otimes B$?
Followup Question (in response to @David Gao's counterexample). What if the range is such that $\dim V_2 =1$?
Background. In physics, we often consider Hermitian operators (called the Hamiltonian $H$) defined on tensor products $\bigotimes_{i\in V}\mathcal{H}_i$ on some set of vertices $V$, consisting of local interactions in the sense that $H=\sum_{e} h_e$ where $h_e$ are Hermitian operator defined on $\mathcal{H}_i\otimes \mathcal{H}_j$ for some $e=ij$ (and thus induces a graph structure on the vertices $V$). I was curious what it meant for each local interaction $h_e$ to be "entangled", i.e., not a trivial product operator.
No. For example, choose a unitary operator $\eta$ on $\mathbb{C}^2 \otimes \mathbb{C}^2$ such that $\eta(e_1 \otimes e_1) = \frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2)$. Since it’s a unitary, its range is $\mathbb{C}^2 \otimes \mathbb{C}^2$. But it cannot be written as $A \otimes B$ as otherwise $\frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2) = \eta(e_1 \otimes e_1) = Ae_1 \otimes Be_1$ is decomposable as a tensor product, which is not possible.
The answer is still no even assuming $\mathrm{dim}(V_2) = 1$. In fact, to settle this completely, the following is an example showing that the answer is no in $\mathbb{C}^2 \otimes \mathbb{C}^2$ (and therefore the answer is no whenever both $\mathcal{H}_1$ and $\mathcal{H}_2$ have dimension at least $2$) even when $\mathrm{dim}(V_1) = \mathrm{dim}(V_2) = 1$. Simply let $\eta$ be defined as the orthogonal projection onto $\mathrm{span}\{\frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2)\}$ composed with the map that sends $\frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2)$ to $e_1 \otimes e_1$. The range is $\mathrm{span}\{e_1\} \otimes \mathrm{span}\{e_1\}$. If $\eta = A \otimes B$, then $\eta^\ast = A^\ast \otimes B^\ast$ and $\eta^\ast (e_1 \otimes e_1) = A^\ast e_1 \otimes B^\ast e_1$, but $\eta^\ast (e_1 \otimes e_1) = \frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2)$.