CatGloss

Following \cite{grothendieck-kansas}, define a fiber space $p \colon E \to B$ to be a morphism in $\textup{\textsf{cat}}$. A map of fiber spaces is a commutative square. Thus, the category of fiber spaces is isomorphic to the diagram category $\textup{\textsf{cat}}^\mathbbe{2}$. We are also interested in the non-full subcategory $\textup{\textsf{cat}}/B \subset \textup{\textsf{cat}}^\mathbbe{2}$ of fiber spaces over $B$ and maps whose codomain component is the identity. Prove the following:

  1. A map \(\xymatrix{ E' \ar[d]_{p'} \ar[r]^g & E \ar[d]^{p} \\ B' \ar[r]_f & B}\)of fiber spaces induces a canonical map from the fiber over a point $b \in B’$ to the fiber over its image $f(b) \in B$.
  2. The fiber of a product of fiber spaces is the product of the fibers. A projection $B \times F \to B$ defines a trivial fiber space over $B$, a definition that makes sense for any space $F$.
  3. Show that the fiber of a trivial fiber space $B \times F \to B$ is isomorphic to $F$.
  4. Characterize the isomorphisms in $\textup{\textsf{cat}}/B$ between two trivial fiber spaces (with a priori distinct fibers) over $B$.
  5. Prove that the assignment of the set of continuous sections of a fiber space over $B$ defines a functor $\textup{fun} \colon \textup{\textsf{cat}}/B \to \textup{\textsf{cat}}$.
  6. Consider the non-full subcategory $\textup{\textsf{cat}}^\2{\text{pb}}$ of fiber spaces in which the morphisms are the pullback squares. Prove that the assignment of the set of continuous sections to a fiber space defines a functor $\textup{fun}\colon (\textup{\textsf{cat}}^\2{\text{pb}})^\mathrm{op} \to \textup{\textsf{cat}}$.
  7. Describe the compatibility between the actions of the ``sections’’ functors just introduced with respect to the map $g$ of fiber spaces $p$ and $q$ over $B$ and their restrictions along $f \colon B’ \to B$. \(\xymatrix@=10pt{ E' \ar@{}[dr]\mid(.2){\displaystyle\lrcorner} \ar[rr] \ar[ddd]_{p'} & & E \ar'[d][ddd]_p \ar[dr]^g \\ & F' \ar@{}[dr]\mid(.2){\displaystyle\lrcorner} \ar[ddl]^{q'} \ar[rr] & & F \ar[ddl]^q \\ & & \\ B' \ar[rr]_f & & B}\)