$Z$ definition? (Hussemoller, fibre bundles)

Question

Hussemoller doesn't describe what $Z$ is in this definition (page 67, third edition):

Note that $\mathbf L(\mathbb F^m, \mathbb F^n)$ is the collection of all linear functions $\mathbb F^m \xrightarrow{\quad} \mathbb F^n$. Also, have $\mathbf{VB}_B$ be the category of vector bundles over a space $B$. Finally, have $\mathbf{VB}_0 (p, q)$ be the product category of $p$ copies of $\mathbf{VB}_B$ and $q$ copies of the respective dual category $\mathbf{VB}_B^*$. Hussemoller does not give a clear definition of $Z$, and uses it as a placeholder for several different objects within the book. The most recent definition I could find is:

There is no clear way to reconcile this definition, since he nondescriptly mentions $z$ as it it were an element of $Z$, without describing the structure or what the map is computationally.

Thanks in advance.

Answer 1

The solution is simple: $Z$ is any space. Husemoller should perhaps better have written

A functor $F: \mathbf{VB}_0(p, q) \to \mathbf{VB}_0$ is called continuous provided for each space $Z$ and each family of maps $u_i : Z \to \mathbf{L}(V_i, W_i)$, where $1 \le i \le p + q$, the function $z \mapsto F(u_1(z), \ldots, u_{p+q}(z))$ is a map $$Z \to \mathbf{L}(F(V_1,\ldots, V_p, W_{p+1},\ldots, W_{p+q}), F(W_j, \ldots, W_p, V_{p+1}, \ldots, V_{p+q}))$$

Recall that a map is understood as a continuous function.

In the proof presented in your question $Z$ is specified to be the sum space of the family $\{V_i \times G\}$.