map<int,int> default values

As soon as you access the map with the [] operator, if the key doesn't exist it gets added. The default initializer of the int type gets invoked - so it will get a value of 0.

Yes, it is safe to assume.

The map's operator[] is specified thus:^{([map.access])}

Effects: If there is no key equivalent to x in the map, inserts value_type(std::move(x), T()) into the map.
Returns: A reference to the mapped_type corresponding to x in *this.

T() uses value-initialisation for all T except void ^{([expr.type.conv]/2)}, and value-initialisation for a primitive results in zero-initialization ^{([dcl.init]/7)}.

Therefore, the expression evaluates to a reference to an object with value zero ^{([dcl.init]/5)}.

The operator++ call then increments that object to one, and evaluates to one.

_{(All references are C++11.)}

Yes, the default value will be the default of that type. If you want another default, you can create a class that behaves like an int but has a different default constructor.