Solving a type of anagram problem with frequency

anagramWithFrequency.js

function isAnagramFunction (word_A, word_B){
        let isAnagram=true;
        let map_A=Object.create(null);
        let map_B=Object.create(null);

        let arr_A = word_A.split('');
        let arr_B = word_B.split('');

        arr_A.forEach(letter=>{
                if(map_A[letter]){
                        map_A[letter]++;
                }else{
                        map_A[letter]=1;
                }
        })
        arr_B.forEach(letter=>{
                if(map_B[letter]){
                        map_B[letter]++;
                }else{
                        map_B[letter]=1;
                }
        })
        Object.keys(map_A).forEach(key=>{
                if(map_B[key]!==map_A[key]){
                        isAnagram=false;
                }
        })
        return isAnagram;
}

console.log(isAnagramFunction("cinema","iceman"));