Suppose I have three sheaves of rings $ A,B,C$ on a topological space $ X$ (or a site), with a map of sheaves of rings $ \gamma: A\otimes_{\mathbf{Z}} B\to C$ .

Suppose I have a continuous self-map of topological spaces $ f : X\to X$ (or a morphism of sites such that $ f^{-1}$ is an exact functor), that comes together with maps of sheaves of rings $ \alpha: f^{-1}A\to A$ and $ \beta: f^{-1}B\to B$ .

Call $ D$ the sub-sheaf of rings of $ C$ on $ X$ generated by the image of the map of sheaves of rings $ \gamma: A\otimes_{\mathbf{Z}} B\to C$ .

Does there exist a map of sheaves of rings $ \delta : f^{-1}D\to D$ that is compatible with $ \alpha$ and $ \beta$ ?

By compatible, I mean that $ \gamma\circ (\alpha\otimes\beta)$ should agree with $ \delta\circ (f^{-1}\gamma)$ , i.e. the obvious diagram of maps of sheaves of rings commutes.

Morally, I’m asking if, given a sheaf of rings $ C$ and two sheaves of subrings $ A,B$ of $ C$ such that $ \alpha$ and $ \beta$ exist, $ \alpha$ and $ \beta$ can be extended to the sheaf of subrings of $ C$ that $ A$ and $ B$ generate.

**Attempt:** For $ U$ open in $ X$ , we call $ D’(U)$ the sheaf of subrings in $ (f^{-1}C)(U)$ generated by the image of $ (f^{-1}\gamma)(U):(f^{-1}A)(U)\otimes_{f^{-1}\mathbf{Z}(U)}(f^{-1}B)(U)\to (f^{-1}C)(U)$ . A section in $ D’(U)$ in the image of $ (f^{-1}\gamma)(U)$ can be sent to a choice of preimage in $ (f^{-1}A)(U)\otimes_{f^{-1}\mathbf{Z}(U)}(f^{-1}B)(U)$ , then to its image in $ A(U)\otimes_{\mathbf{Z}(U)}B(U)$ via $ \alpha(U)\otimes\beta(U)$ , and then to $ D(U)$ . This assignment is well defined because $ f^{-1}$ commutes with equalizers, and then sends the kernel of $ (f^{-1}\gamma)(U)$ to the kernel of $ \gamma(U)$ . It is also additive and multiplicative, and hence uniquely extends to a ring homomorphism $ \delta(U) : D’(U)\to D(U)$ .

One should be able to check by hand that $ \delta(U)$ is natural in $ U$ , and gives a morphism of sheaves of rings $ D’\to D$ .

The question becomes whether $ f^{-1}D = D’$ as sheaves of rings. This sounds plausible but I’m not sure how to see it.