what's .map() doing?

I found an interesting question on StackOverflow, saying:
Using console in Chrome, input is

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[0].map(Array);

and output is

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[0, 0, [0]];

It looks confused at first, and by looking up the reference on MDN, I found

var new_array = arr.map(function callback(currentValue, index, array) {

}[, thisArg])

and that’s easily explained.
We pass currentValue = 0, index = 0, array=[0] into Array: we got

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Array(0, 0, [0]);

Wait, how about multiple values?

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[0, 0].map(Array);

we got

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2
3
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[
    [0, 0, [0, 0]],
    [0, 1, [0, 0]],
];

We pass currentValue = 0, index = 0, array=[0] into Array, and then, we pass currentValue = 0, index = 1, array=[0] into Array. And map returns an array.

no problem!

How about Number?

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[0].map(Number);

should that return a

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[0, 0, 0];

No way. And why?
should it pass currentValue = 0, index = 0, array=[0] into Number?
It should, but Number only takes one argument, the first one. It equals that we pass 0 into Number:

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Number(0);

It seems that we solved full questions. But that’s not the end.

parseInt shows different result

trying:

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[0, 0].map(parseInt);

You must think we got [0, 0]
Some clever boy looks up reference on MDN:

parseInt(string, radix);

it equals parseInt([0, 0])
Oh, we must got [0] instead.
Well, how about ?

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[10, 10].map(parseInt);

emmmm, We will got [10], because 10 in 10-radis is 10
Holy wrong!
Remember, we pass currentValue, index, array for each value?
index is important!
The answer is left for you.

What do you get from this passage?

Reference