Assume you have an unknown word on an alphabet with at least three letters, and you know all the words obtained by erasing each copy of some letter. Then, you can find the first letter of the original word (it is the only one satisfying the following property : it is at the beginning of every word obtained by removing a letter which is not itself). By continuing that way, you recover the whole word. So the map from the free monoid on $n$ letters to the product of $n$ free monoid on $n-1$ letters obtained by erasing all copies of some letters is injective.
Now I'm asking the same question for free groups. Of course, a word in the kernel is trivial in the abelian free group (the algebraic total number of occurrences of some letter can be recovered by removing any other letter), so it is a product of commutators. Also, it is (tautologically) for every letter $l$ a product of conjugates of $l$. But is it necessarily an empty word?
Geometrically : if a loop on the disk with $n$ points removed ($n\geq 3$) is nullhomotopic each time you add one the points back, is it necessarily nullhomotopic?
I'm afraid the answer to these questions is no, but I'm failing to either find a proof, a counterexample, or a reference addressing this question. I'm also thinking maybe $n\geq 4$ or something could work with $3$ failing.
What do you folks think? Thanks for any help :)