My professor had limited to talk about matrices, then since the functional calculus for self-adjoint matrices $M \in \mathbb{M}_{n\times n}$ is such that $\sigma(f(M))=f(\sigma(M))$ he then defined:
Definition: We define the observable presheaf $\mathcal{S}:\mathbb{M}_{n\times n} \rightarrow \mathbf{Set}$ as the contravariant functor that maps $ \mathbb{M}_{n\times n} \ni M \stackrel{\mathcal{S}}{\mapsto} \sigma(M) \in \text{Obj}_\mathbf{Set}$ and for the morphisms from a element $A$ to a element $B$ of $\mathbb{M}_{n \times n}$, taken to be the functional calculus $\Phi_B(f) \equiv f(B)$ whenever $A=f(B)$, $\,\mathcal{S}$ is such that $\mathcal{S}(\Phi_B(f))=f:\sigma(B) \rightarrow \sigma(A)$.
But for more general observables than matrices, the Borel functional calaculus is such that only $\sigma(f(A)) \subset \overline{f(\sigma(A))}$ is true in general, there was some comment about it requiring more technical sophistication to be able to formulate the observable presheaf for arbitrary $C^*$-algebras having to do with dealing with topology, and I believe compact sets, which somewhat makes sense to me since there is a note on the book about spectral theory that I am reading that the reverse containment, i.e. $f(\sigma(A)) \subset \sigma(f(A))$, is not true in general since bounded borel functions do not take compact sets to compact sets necessarily.
I had searched about this, but the things that are usually called "observable presheaf" do not map objects $A$ to $\sigma(A)$, like for example the $\mathcal{O}^+_\mathcal{R}$ (or $\mathcal{O}^-_\mathcal{R}$) in https://browse.arxiv.org/pdf/0708.0677.pdf which sends abelian von Neumann subalgebras of a von Neumann algebra $\mathcal{R}$, to the self-adjoint part of the abelian von Neumann subalgebra, in symbols: $\mathfrak{A}(\mathcal{R}) \ni \mathcal{A} \stackrel{\mathcal{O}^{+(-)}_\mathcal{R}}{\mapsto} \mathcal{A}_{sa}$ (which is a presheaf since presheafs need to be contravariant functors arriving at the category $\mathbf{Set}$, and $\mathcal{A}_{sa}$ can be faced as a set, but not as a subset of $\mathbb{R}$ which would be necessary for me).