Q:

Fibrewise product

I have recently started studying fibrewise topology.
It is not clear to me what is the difference between the normal product space and the fibrewise product space over a topological space B.
I am following James book and it defines it as a subset of the normal product space.
Thanks for your help

A:

Let $X,Y$ be topological spaces. Then, of course, their product $X\times Y$ is the cartesian product of the underlying sets $U(X),\ U(Y)$ of $X$ and $Y$ respectively, equipped with the coarsest topology making both projections of $U(X)\times U(Y)$ onto the factors continuous. With this definition, $X\times Y$, together with the projections $p_{X},\ p_{Y}$ satisfies the following universal property: given any pair of continuous functions $(h\colon Z\to X,\ k\colon Z\to Y)$, there exists a unique continuous map $(h,\ k)\colon Z\to X\times Y$ such that $p_{X}\circ (h,k)=h$ and $p_{Y}\circ (h,k)=k$. (The proof of this claim is an easy, but mandatory, exercise).
Now, suppose to have a couple of continuous functions $f\colon X\to Z$ and $g\colon Y\to Z$ into the same topological space $Z$. Then you know the definition of $X\times_{Z}Y$ for this pair of map as the set
$$U(X\times_{Z}Y):=\{(x,y)\in U(X)\times U(Y):\ f(x)=g(y)\}$$
endowed with the subspace topology inherited by the product topology given on $X\times Y$. Note that $X\times_{Z}Y$ comes together with two continuous map $\pi_{X}\colon X\times_{Z}Y\to X$ and $\pi_{Y}\colon X\times_{Z}Y\to Y$, which are the restrictions of $p_{X}$ and $p_{Y}$ respectively. The triple $(X\times_{Z}Y,\ \pi_{X},\ \pi_{Y})$ is such that $f\circ \pi_{X}=g\circ \pi_{Y}$ and satisfies the following unviversal property: given any triple $(W,\ h\colon W\to X,\ k\colon W\to Y)$ such that $f\circ h=g\circ k$, there is a unique continuous map $t\colon W\to X\times_{Z}Y$ such that $p_{X}\circ t=h$ and $p_{Y}\circ t=k$. (Again, prove this!)
It follows, that in general the two concept of product and fibered product are distinct, that is, $X\times Y$ is not homeomorphic to $X\times_{Z}Y$. (As, more generally, a subspace of a topological space is not homeomorphic to the whole topological space, of course). For example, consider a continuous map $f\colon X\to Y$ and the identity map $1_{Y}\colon Y\to Y$. Then the fibered product $X\times_{Y} Y$ with rispect to these maps is  just the graph of $f$ (look at the definitions!), which, of course, is not homemorphic to $X\times Y$ in general.
Note however that, if you take the only possible (continuous) map $f\colon X\to\{0\}$ and $g\colon Y\to\{0\}$, you can see that $X\times_{\{0\}} Y=X\times Y$. So, if you want, fibered products generalise products, as the latter can be recovered as particular instances of the former.
Hope this helps somehow.

