Vector Bundles: Continuity of map between total space implies homeomorphism. In Spivak's "A Comprehensive Introduction to Differential Geometry" Spivak defines a vector bundle as a tuple: $(E, \pi, B, \bigoplus, \bigodot),$ where $E$ is the total space, $B$ is the base space, $\pi$ is a continuous map from $E$ onto $B$ (thought of as like a projection) and there are vector additions on each fiber, and scalar multiplication on the total space. Lastly, he gives a local trivialization condition.
He says that two vector bundles over the same space are equivalent if there is a homeomorphism of the total spaces which takes fibers onto fibers. This is what he says is the definition of equivalence. 
In an exercise after this, he says that we do not need a homeomorphism - that if the map between total spaces is continuous, then the spaces are in fact homeomorphic.
I don't even begin to see what's going on here. For manifolds in general, this is far from true - there is no reason continuity should say anything about the inverse map being continuous, and certainly not bijectivity. The only thing I can think of is that Spivak's question is worded a little poorly, and I should assume the map is continuous, and the fibers are taken isomorphically to fibers.  Even with this new information though, I still feel mostly lost.
For example, why can't we send the whole total space to just a single point and fiber? 
Thanks for any pointers or hints.
 A: Though the statement you mention above is somewhat loosely phrased let us try to understand what it means. 
Notation: Our definition of a vector bundle is a tuple $(E,B, \pi)$ (I am ignoring the other components in your definition as they aren't relevant to us.) By local triviality I can write down points in the total space $E$ with co-ordinates $(b,t)$ where $b \in B$ and $t \in F_{b}$ where $F_{b}$ is the fiber over $b$.
We consider the case of two bundles $E, E^{\prime}$ and a map of the total spaces $f: E \rightarrow E^{\prime}$ realizing this equivalence. The underlying base space is $B$. 
In this case the equivalence is described by :
Definition 1
"He says that two vector bundles over the same space are equivalent if there is a homeomorphism of the total spaces which takes fibers onto fibers. This is what he says is the definition of equivalence.
"
Consider the following alternate definition for equivalence of vector bundles. 
Definition 2  An equivalence of vector bundles is an arrow $\varphi : (E, B, \pi) \rightarrow (E^{\prime}, B, \pi^{\prime})$  which induces (1) a continuous map $\varphi: E \rightarrow E^{\prime}$ that is an isomorphism (vector space) on the fibers and (2) the induced maps from $E$ to $B$,  $\pi^{\prime} \varphi$ and $\pi$ are equal. 
If we were given a $\varphi$ as above, let $\varphi_{b}$ denote the induced linear isomorphism between the fibers $F_{b}$ and $F^{\prime}_{b}$ over the point $b \in B$. Consider the function $\varphi^{-1}: E^{\prime} \rightarrow E$ defined at a point $(b,t) \in E^{\prime}$ as follows:
 $$\varphi^{-1}(b,t) = (b, \varphi_{b}^{-1}(t)).$$
we claim that $\varphi^{-1}$ is a continuous map of topological spaces- this follows again by local triviality. Hence $\varphi^{-1}$ defines an inverse of vector bundle map $\varphi$ according to definition 2. 
Let us see the equality of these definitions. 
$(2 \implies 1)$
By local triviality any point $e\in E$ has a co-ordinate $(b,t)$ where $b\in B$ and $t$ is in the fiber. If we were given a map $\varphi$ according to definition (2) we define $f(b,t) = (b, \varphi_{b}(t))$. This makes $f$ into a continuous and bijective map. We consider now the map $g: E^{\prime} \rightarrow E$ by $g(b,t) = (b, \varphi^{-1}_{b}(t))$. By the same argument $g$ is a continuous bijective map and hence it makes $f$ into a homeomorphism. 
$(1 \implies 2)$
In this case we define $\varphi(b,t) = f(b,t)$. In this case everything is simpler except perhaps the condition $\pi^{\prime} \varphi = \pi$. To this end, given a vector bundle $(E,B, \pi)$ we have a continuous map of topological spaces $z: B \rightarrow E$ given by $z(b) = (b,0)$. If you aren't sure check this is continuous -local triviality. 
So the map composite map of topological spaces $$ B \stackrel{z}{\rightarrow} E \stackrel{f}{\rightarrow} E^{\prime} \stackrel{\pi^{\prime}}{\rightarrow} B$$
is identity and so is $\pi\circ z: B \rightarrow B$. This gives the required equality.
