Line bundle trivial on fibers then isomorphic to the pullback of a line bundle $\require{AMScd}$
I'm currently reading Milne's notes about Abelian varieties. On page 26 he proves the following theorem:

Let $V$ and $T$ be varieties over $k$ with $V$ complete, and let $\mathcal{L}$ be an invertible sheaf on $V\times_k T$. If $\mathcal{L}_t$ is trivial for all $t \in T$,then there exists an invertible sheaf $\mathcal{N}$ on $\mathcal{L}$ such that $\mathcal{L}=q^* \mathcal{N}$ 
  $
\begin{CD}
V\times T @<i<< V_t \\
@VVqV @VVpV\\
T @<j<< t
\end{CD}$

Here Milne denotes by $\mathcal{L}_t$ the pullback of $\mathcal{L}$ along the map $i$.
To prove this theorem he shows first that $q_* \mathcal{L}= \mathcal{N}$ is an invertible sheaf. Now to complete the proof we want to show that the natural map $\alpha : q^* \mathcal{N}=q^* q_* \mathcal{L} \rightarrow \mathcal{L}$ is an isomorphism. 

If we restrict $\alpha$ to a fibre $V_t$ Milne claims that we get the
   following map $\alpha_t :  O_{V_t}\otimes_{k(t)} \Gamma (V_t, O_{V_t})
 \rightarrow \mathcal{L}_t$. Why is this true?

. If I unwrap $q^* q_* \mathcal{L}|V_t$ I get that it is equal to $$i^*q^* \mathcal{N}=i^*(q^{-1}\mathcal{N}\otimes_{q^{-1}O_T} O_{V\times T})$$
Now it is clear that $i^*O_{V\times T}=O_{V_t}$. But why is $i^*q^{-1}\mathcal{N}= \Gamma (V_t, O_{V_t})$ and $i^*q^{-1}O_T=k(t)$?
Now $\alpha_t$ is an isomorphism which for all $w\in V_t$ induces an isomorphism $ \alpha_{t,w} : ((qi)^* \mathcal{N})_w \rightarrow \mathcal{(i^*L)_w}$. My second question is: 

how does this induces an isomorphism $$ \alpha(w) : \frac{q^*
 \mathcal{N}_w}{\mathcal{m}_w q^* \mathcal{N}_w} \rightarrow
 \frac{\mathcal{L}_w}{\mathcal{m}_w \mathcal{L}_w}$$

Thanks for any help in advance!
 A: $\newcommand{\sheaf}[1]{{\mathcal{#1}}}$
$\newcommand{\Ohol}{{\mathcal{O}}}$
This is exactly a problem I also had to think much about, when studying the semicontinuity theorems in Hartshone's AG. I would argue as follows
Start with $q^* q_* \sheaf{L} \to \sheaf{L}$ and apply $i^*$ (restrict to fiber over $t$) to get
$i^* q^* q_* \sheaf{L} \to i^* \sheaf{L} = \sheaf{L}_t$.
Now, using $i^* \circ q^* = p^* \circ j^*$ this is
$$p^* j^* q_* \sheaf{L} \to \sheaf{L}_t$$
Now use the relation 
$$(*) \quad j^* q_* \sheaf{L} = p_* i^*\sheaf{L}$$
to get
$$p^* p_* i^*\sheaf{L} \to i^*\sheaf{L}$$
that is 
$$p^* p_* \sheaf{L}_t \to \sheaf{L}_t$$
The relation $(*)$ is nothing else but Corollary 12.9 in Hartshorne's AG,
written for $i=0$ (note $i$ is an index in AG, not the map used above) and $f = q$.
This is exactly the map
$$\alpha_t: \Ohol_{V_t} \otimes_{k(t)} H^0(V_t,\sheaf{L}_t) \to \sheaf{L}_t$$
By replacing $\Gamma(V_t,\Ohol_{V_t})$ with the (isomorphic but) functorially correct  $H^0(V_t,\sheaf{L}_t)$ it becomes clear, how $\alpha_t$ is defined.
As for a point $w$ in $V_t$ we have $\Ohol_{V_t,w} \cong {\sheaf{L}_t}_w$ it follows for $a:\{w\} \to V_t$ that $a^*(\alpha_t)$ is an isomorphism for all $w \in V_t$. So by Nakayama's Lemma $\alpha_t$ is an isomorphism of it's arguments. Rewinding the above equalities we get
$$a^*i^*q^*q_* \sheaf{L} \to a^*i^*\sheaf{L}$$
is an isomorphism for all $a:\{w\} \to V_t$ as above.
So $q^*q_*\sheaf{L} \to \sheaf{L}$ is an isomorphism.
