Fundamental theorem of topos theory

In mathematics, The fundamental theorem of topos theory states that the slice $$\mathbf{E} / X$$ of a topos $$\mathbf{E}$$ over any one of its objects $$X$$ is itself a topos. Moreover, if there is a morphism $$f : A \rightarrow B $$ in $$\mathbf{E}$$ then there is a functor $$f^*: \mathbf{E} / B \rightarrow \mathbf{E} / A $$ which preserves exponentials and the subobject classifier.

The pullback functor
For any morphism f in $$\mathbf{E}$$ there is an associated "pullback functor" $$ f^* := - \mapsto f \times - \rightarrow f $$ which is key in the proof of the theorem. For any other morphism g in $$\mathbf{E}$$ which shares the same codomain as f, their product $$f \times g$$ is the diagonal of their pullback square, and the morphism which goes from the domain of $$f \times g$$ to the domain of f is opposite to g in the pullback square, so it is the pullback of g along f, which can be denoted as $$f^*g$$.

Note that a topos $$\mathbf{E}$$ is isomorphic to the slice over its own terminal object, i.e. $$\mathbf{E} \cong \mathbf{E} / 1$$, so for any object A in $$\mathbf{E}$$ there is a morphism $$f : A \rightarrow 1 $$ and thereby a pullback functor $$f^* : \mathbf{E} \rightarrow \mathbf{E} / A$$, which is why any slice $$\mathbf{E} / A $$ is also a topos.

For a given slice $$\mathbf{E} / B$$ let $$X \over B$$ denote an object of it, where X is an object of the base category. Then $$B^*$$ is a functor which maps: $$ - \mapsto {B \times - \over B} $$. Now apply $$f^*$$ to $$B^*$$. This yields
 * $$f^* B^* : - \mapsto {B \times - \over B} \mapsto {{A \over B} \times {B \times - \over B} \over {A \over B}} \cong {\Big({A \times_B \, B \times - \over B}\Big) \over {A \over B}} \cong {A \times_B B \times - \over A} \cong {A \times - \over A} = A^*$$

so this is how the pullback functor $$f^*$$ maps objects of $$\mathbf{E} / B$$ to $$\mathbf{E} / A$$. Furthermore, note that any element C of the base topos is isomorphic to $${1 \times C \over 1} = 1^* C$$, therefore if $$f : A \rightarrow 1$$ then $$f^*: 1^* \rightarrow A^*$$ and $$f^* : C \mapsto A^* C $$ so that $$f^*$$ is indeed a functor from the base topos $$\mathbf{E}$$ to its slice $$\mathbf{E} / A$$.

Logical interpretation
Consider a pair of ground formulas $$\phi$$ and $$\psi$$ whose extensions $$[\_|\phi]$$ and $$[\_|\psi]$$ (where the underscore here denotes the null context) are objects of the base topos. Then $$\phi$$ implies $$\psi$$ if and only if there is a monic from $$[\_|\phi]$$ to $$[\_|\psi]$$. If these are the case then, by theorem, the formula $$\psi$$ is true in the slice $$\mathbf{E} / [\_|\phi]$$, because the terminal object $$[\_|\phi] \over [\_|\phi]$$ of the slice factors through its extension $$[\_|\psi]$$. In logical terms, this could be expressed as
 * $$\vdash \phi \rightarrow \psi \over \phi \vdash \psi$$

so that slicing $$\mathbf{E}$$ by the extension of $$\phi$$ would correspond to assuming $$\phi$$ as a hypothesis. Then the theorem would say that making a logical assumption does not change the rules of topos logic.