How to prove a bundle map is a bundle isomorphism? Example #1
In proving pullback bundle is homotopy invariant Ralph Cohen's notes on the topology of fiber bundles use the following proof based on the covering homotopy theorem (pp.47):
Let $p: E \to B$ be a fiber bundle and $f_0, f_1: X \to B$ be homotopic to each other with the homotopy given by $H: X \times I \to B$.  By the covering homotopy theorem we have the following fiber bundle over $X \times I$.
$\require{AMScd}$
\begin{CD}
  f_0^*(E) \times I @>\tilde{H}>> E \\
  @VVV @VVpV \\
  X \times I @>H>> B
\end{CD}
Then he claims that by definition this defines a map of bundles over $X \times I$
$\require{AMScd}$
\begin{CD}
  f_0^*(E) \times I @>>> H^*(E) \\
  @VVV @VVV \\
  X \times I @= X \times I
\end{CD}
which is clearly a bundle isomorphism since it induces the identity map on both the base space and on the fibers.
I don't understand why $f_0^*(E) \times I \to H^*(E)$ is an isomorphism.  I agree that it induces the identity on the base space but how can the map on fibers is also the identity?  If both are identity wouldn't that imply the two total spaces are the same?
Example #2
On the same notes pp.53 during the proof of the bijection between $[X,B]$ and isomorphism classes of principal bundles over $X$ where $(h,f)$ defines a homeomorphism of principal bundle as below
$\require{AMScd}$
\begin{CD}
  P @>h>> E \\
  @VqVV @VVpV \\
  X @>f>> B
\end{CD}
which induces an equivariant isomorphism on each fiber thus induces the following bundle isomorphism to the pullback
$\require{AMScd}$
\begin{CD}
  P @>>> f^*(E) \\
  @VqVV @VVpV \\
  X @= X
\end{CD}
again I don't understand why $P \to f^*(E)$ is a bundle isomorphism.
Example #3
On the same notes pp.48 in showing K-theory is a homotopy invariant, the missing step is to show if $E \to Y$ and $E' \to Y$ are isomorphic bundles over $Y$ i.e.
$\require{AMScd}$
\begin{CD}
  E @>\cong>> E' \\
  @VVV @VVV \\
  Y @>\cong>> Y
\end{CD}
then their pullback bundles along the same map $f: X \to Y$ is also isomorphic i.e.
$\require{AMScd}$
\begin{CD}
  f^*(E) @>\cong>> f^*(E') \\
  @VVV @VVV \\
  X @>\cong>> X
\end{CD}
The global homeomorphism between $E$ and $E'$ induces homeomorphism on their local trivializations which defines a homeomorphism on local trivializations of $f^*(E)$ and $f^*(E')$ (as the local trivialization of the pullback is defined using the local trivialization of the bundle).  Global homeomorphism between $f^*(E)$ and $f^*(E')$ then follows from the pasting lemma?
 A: In the first case, the explanation given in the notes is indeed rather poor.  The fact that the map $f_0^*(E) \times I \to H^*(E)$ is an isomorphism follows not from the statement of the covering homotopy theorem (which, as those notes state it, doesn't even guarantee that $\tilde{H}$ induces homeomorphisms on each fiber) but from its proof.  When you examine the proof, you can see that it constructs a certain bundle map $f_0^*(E) \times I \to H^*(E)$ which is seen to be continuous by the pasting lemma (it is constructed by pasting together pieces which lie over a cover of $B\times I$ by certain closed sets $A_{ij}=\bar{W}_i'\times [t_j,t_{j+1}]$).  But actually, over each of these pieces $A_{ij}$, there are trivializations of both $f_0^*(E)\times I$ and $H^*(E)$ given by $\phi_k$ (since each of them is pulled back from a bundle on $U_\alpha$ which is trivialized by $\phi_k$), and when you use these trivializations to identify both bundles with the trivial bundle $A_{ij}\times F$, the map is just the identity map.  This is what the notes mean when they say the map is "the identity on fibers": when you choose these local trivializations induced by $\phi_k$ on each piece of the base, the map becomes the identity on fibers.  It follows that the inverse map, similarly defined piecewise, is also continuous, so the map is a homeomorphism.
In the second case, the explanation is much simpler: any equivariant map between principal bundles which covers the identity map is automatically a homeomorphism.  To see this, you can look locally on the base, so you can assume both your bundles are trivial.  Then you have map $\alpha:X\times G\to X\times G$ of the form $\alpha(x,g)=(x,\beta(x)g)$ for some continuous map $\beta:X\to G$ (the map $\beta$ is continuous because it is the second coordinate of the map $\alpha(x,1)$).  The inverse of $\alpha$ is then just $\alpha^{-1}(x,g)=(x,\beta(x)^{-1}g)$, which is continuous because the inverse map $G\to G$ is continuous.
