What is Thom Isomorphism? I am reading the following post on Thom Isomoprhism and I also have the Thom Isomoprhism from Hatcher's, Corollary 4.9,pg441
nlab's: Let $V \rightarrow X$ be a rank $n$ vector bundle over a simply connected connected $CW$ complex $B$. Let $R$ be a commutative ring. There exists an element $c \in H^n(Th(V);R)$ such that forming the cup product with $c$ induces an isomoprhism $$H^*(B;R) \rightarrow \tilde{H}^{*+n}(Th(V);R), x \mapsto c \cup  \pi^*x $$
Hatcher's: if the disk bundle $(D^n, S^{n-1}) \rightarrow (E,E')\xrightarrow{p} B$ has a Thom class $c \in H^n(E,E';R)$ then the map $\Phi:H^i(B, R) \rightarrow H^{i+n}(E,E';R), \Phi(b)=p^*b \cup c$ is an isomorphism for all $i \ge 0$ and $H^i(E,E';R) =0$ for $i<n$.

It doesn't seem clear to me. But I suppsoe nlab's are always in the most general case. How does nlab's imply Hatcher's? Or does one imply the other?

Note: A Thom class in Hatcher's definition is an element $c \in H^n(E,E';R)$ such that when restricted to each fiber $(D^n,S^{n-1})$is a generator.
 A: The CW complex $X$ is paracompact (see Hatcher for the reference), so in particular admits a locally finite partition of unity. Using this object one can put a bundle metric on any (locally trivial) vector bundle $p:V\rightarrow X$ of finite rank $n$ by piecing together the standard bundle metrics that exist over each local trivialisation (I'm sure Hatcher's notes on vector bundles must cover this).
Now that a $V$ is endowed with a bundle metric $g$ you can construct the associated disc bundle
$$D(V)=\{e\in V\mid g(e,e)\leq 1\}$$
and the associated sphere bundle
$$S(V)=\{e\in V\mid g(e,e)=1\}.$$
The projection $p$ restricts to each of these spaces to give maps to $X$, and the local trivialisations of $V$ restrict to turn $D(V)\rightarrow X$ and $S(V)\rightarrow X$ into locally-trivial fibre bundles with fibres $D^n$ and $S^{n-1}$, respectively. Note that $S(V)\subseteq D(V)$ by construction.
The Thom Space of $V$ is now defined as the quotient space of the disc bundle by sphere bundle
$$Th(V)=D(V)/S(V).$$
The notation that Hatcher is using is $E=D(V)$, $E'=S(V)$.
Hence there is long-exact sequence in cohomology
$$\dots\rightarrow H^nD(V)\rightarrow H^nS(V)\rightarrow H^n(D(V),S(V))\rightarrow \dots$$
where $H^n(E,E')=H^n(D(V),S(V))\cong H^n(D(V)/S(V))=H^n(Th(V))$. 
The nlab page is working under the assumption that $X$ is simply connected. Therefore let us choose a basepoint $x\in X$ and an isomorphism $H^n(D(V)_x,S(V)_x)\cong H^n(D^n,S^{n-1})\cong H^n(S^n)\cong\mathbb{Z}$ of the relative cohomology of the fibres over $x$ to get a generator $t_x\in H^n(D(V)_x,S(V)_x)$. This basis element now transports uniquely to a generator $t_y\in H^n(D(V)_y,S(V)_y)\cong\mathbb{Z}$ for the cohomology of the fibre over any other point $y$ contained in a suitable local bundle chart $U$ at $x$. In this way we get a generator $t_U$ for $H^n(D(V)|_U,S(V)|_U)\cong\mathbb{Z}$.
If $U'$ is another bundle chart for $V$ which intersects $U$ non-trivially then we construct a second generator $t_{U'}\in H^n(D(V)|_{U'},S(V)|_{U'})\cong\mathbb{Z}$. Using a Mayer-Vietoris argument we can arrange to choose $t_{U'}$ so that there is agreement $t_U|=t_{U'}|\in H^n(D(V)|_{U\cap U'},S(V)|_{U\cap U'})$.
Continuing in this way we arrive at a global class $t\in H^n(D(V),S(V))$ which for any point $x\in X$ restricts to a generator of $H^n(D(V)_x,S(V)_x)$. That is, $t$ is exactly a Thom class for $V$.
Hatcher's more general statement now follows.
