The order of ε-transitions in NFA

I'm reading 'Introduction to the Theory of Computation' by Michael Sipser. He gives an example of a NFA, stating that this particular automaton accepts $$a$$ (he lists other strings as well, I just want to focus on $$a$$).

But when he demonstrates how to convert the NFA to an equivalent DFA, he says: state $$1$$ goes to $$\emptyset$$ on $$a$$ because no $$a$$ arrows exit it. [...] Note that the procedure in Theorem 1.39 [on equivalence of DFAs and NFAs] specifies that we follow the $$\epsilon$$ arrows after each input is read.

So, if $$\epsilon$$-transitions happens only after some input (as the author defines), then in the initial state $$1$$, when $$a$$ is given, the NFA should 'die' as $$\delta(1, a) = \emptyset$$. The transition to state $$3$$ never happens, the NFA doesn't accept $$a$$. Doesn't the author contradicts himself?

I can think of couple possible explanations: 1) the very first transition of a NFA is $$\delta(1, \epsilon)$$ just by definition, 2) $$a\epsilon=a$$, so $$\delta(1, a)$$ = $$\hat{\delta}(1, a\epsilon)$$ = $$\delta(1, \epsilon)\cup\delta(1, a)$$, but then again the author said state $$\{1\}$$ goes to $$\emptyset$$ on $$a$$, not $${3}$$

I think the answer is somewhat subtle and lies in Sipser's definition of what it means for an NFA to accept a string.

Let $$N = (Q,\Sigma,\delta,q_0,F)$$ be an NFA, and $$w$$ a string over the alphabet $$\Sigma$$. Then we say that $$N$$ accepts $$w$$ if we can write $$w$$ as $$w=y_1y_2\dots y_m$$, where each $$y_i$$ is a member of $$\Sigma_\varepsilon$$ and ...

Compare this with his definition of what it means for a DFA to accept a string:

Let $$M = (Q,\Sigma,\delta,q_0,F)$$ be a DFA and let $$w=w_1w_2\dots w_n$$ be a string where each $$w_i$$ is a member of the alphabet $$\Sigma$$. Then ...

Therefore, by his definitions, his NFA $$N_4$$ accepts $$a$$ as it can be written as $$a=\varepsilon a$$, and $$\varepsilon$$ and $$a$$ are both members of $$\Sigma_\varepsilon$$.

An NFA can always choose to take an $$\varepsilon$$-transition (if one exists) without reading any input, so your NFA would indeed accept the string $$a$$ by following the transitions $$1 \xrightarrow{\varepsilon} 3 \xrightarrow{a} 1$$ and ending up in an accepting state. This is because the NFA nondeterministically chose to follow the $$\varepsilon$$-transition before reading $$a$$ from the input.

However, when converting an NFA to a DFA we have to resolve the nondeterminism in some way. The procedure presented by Sipser in Theorem $$1.39$$ follows the convention that $$\varepsilon$$-transitions are followed after each input symbol is read, but, as he also states, a procedure based on following $$\varepsilon$$-transitions before each input symbol is read would work equally well. It is only a matter of the convention used to resolve the nondeterminism in the NFA.

If you look at Example $$1.41$$ and the DFA resulting from the conversion in Figure $$1.43$$, you will notice that the initial state is $$\{1,3\}$$ (the union of states $$1$$ and $$3$$ from the NFA) because the NFA have initial states $$1$$, but it can also go to $$3$$ with an $$\varepsilon$$-transition without reading any input. Therefore, the initial state in the DFA is the union of the two.