Using Enum.reduce ultimately delegates to the protocol function Enumerable.reduce, which is implemented for Map here:

This first converts the Map to a list of {key, value} tuples, then reduces over it.

Tangentially related, I love the StackOverflow Axiom at play here: “It’s easier to get a good answer by proposing an incorrect implementation than by asking a well-stated question.”

That won’t work, there is no += in elixir. I think what you really mean is more like this:

alphabet_positions = %{"a" => 1, "b" => 2, "c" => 3, …}
Enum.reduce(alphabet_positions, 0, fn {letter, position}, sum ->
  if letter in ~w[a e i o u], do: sum + position, else: sum
end)

No, its turned into a list of tuples with the key as the first element and the corresponding values as the second. The implementation of Map.to_list/1 is equivalent to this:

map
|> Map.keys()
|> Enum.reduce([], fn key, list -> [{key, map[key]} | list] end)

Going the other way (list to map) Map.new/2 can be helpful.

iex(1)> square_tuple = fn x -> {x , x * x} end
#Function<7.91303403/1 in :erl_eval.expr/5>
iex(2)> result = Map.new(1..10, square_tuple)
%{
  1 => 1,
  2 => 4,
  3 => 9,
  4 => 16,
  5 => 25,
  6 => 36,
  7 => 49,
  8 => 64,
  9 => 81,
  10 => 100
}
iex(3)>