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Let $A,B$ be monoids and $A\amalg B$ their product in the category of monoids, comprised of reduced words. Previously I have asked about the canonical arrow $A\amalg B\to A\times B$ given by e.g $$a_1a_2b_1a_3b_2b_3b_2a_2\mapsto(a_1a_2a_3a_2,b_1b_2b_3b_2).$$ This question is further along those lines. In catgories like unital mamgas and monoids the mentioned canonical arrow is always surjective. I am wondering about its injectivity.

  1. The fiber of $(a,b)$ contains both $ab,ba$. Thus for this arrow to be injective these must coincide i.e $a,b$ commute in $A\amalg B$. Shouldn't there be no relations between elements of $A$ and $B$ in the coproduct? Does the equality $ab=ba$ imply $a$ or $b$ are the unit of $A$ or $B$? In particular, does the injectivity of the canonical arrow imply $A$ or $B$ is zero (trivial)?

  2. The fiber of $(a_1a_2,b)$ contains for instance both $a_1ba_2$ and also $a_2ba_1$. Suppose these are equal. Since there are no relations between $A,B$ in the coproduct, can we cancel the $b$ and conclude $a_1a_2=a_2a_1$?

  3. Does the equality $a_1b=a_2b$ imply $a_1=a_2$?
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  • $\begingroup$ The claim in the first sentence of 2 is not right. Did you mean to write something else? $\endgroup$ Nov 5 '16 at 21:20
  • $\begingroup$ @AlexKruckman yes! I edited. $\endgroup$
    – Arrow
    Nov 6 '16 at 8:17
  • $\begingroup$ (1) Nontrivial $a\in A,b\in B$ do not commute in $A\sqcup B$. (3) Yes. (2) $a_1ba_2=a_2ba_1$ is only possible if $a_1=a_2$. $\endgroup$ Nov 6 '16 at 8:34
  • $\begingroup$ @arctictern great. How can I prove these facts? $\endgroup$
    – Arrow
    Nov 6 '16 at 8:44
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    $\begingroup$ You can explicitly construct $A\sqcup B$ to be the monoid of all words in $A$ and $B$ where the only simplifications occur are from inside $A$ and $B$. Prove it satisfies the universal property. $\endgroup$ Nov 6 '16 at 8:51
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All of these questions can easily be answered by explicitly constructing $A\coprod B$ as a set of "reduced words". Specifically, let a reduced word on $A$ and $B$ be a finite sequence of elements of the disjoint union of $A\setminus\{1\}$ and $B\setminus\{1\}$ where you alternate between taking elements of $A$ and elements of $B$. To multiply two words, you concatenate them, and then "reduce", meaning that if concatenating has made two consecutive letters of your word come from the same monoid, you compose them in that monoid. If the composition is the identity, you then remove it, and continue reducing in the same way until you can't anymore.

This operation makes the set of reduced words a monoid with natural inclusion maps from $A$ and $B$ (sending non-identity elements to words of length $1$ and the identity to the empty word). It is then straightforward to check that this monoid is a coproduct of $A$ and $B$. (The universal property is basically immediate from the fact that it is generated by $A$ and $B$ and the definition of the multiplication only uses relations that are forced by the monoid axioms. The hardest part about checking that this construction actually works is checking that the "reduced concatenation" operation is associative, so you really do have a monoid.)

So to answer your questions:

  1. If $a$ and $b$ are both not the identity, then $ab$ and $ba$ are both reduced words, so they are distinct elements of $A\coprod B$.
  2. If $b$ is not the identity and $a_1\neq a_2$, then $a_1ba_2$ and $a_2ba_1$ are distinct elements of $A\coprod B$ (if $a_1$ and $a_2$ are not the identity they are two distinct reduced words; if one of them is the identity this reduces back to question 1). So it's not so much that we can cancel $b$, as that $a_1ba_2=a_2ba_1$ can't be true to begin with unless $b=1$.
  3. Again, if $a_1\neq a_2$, it is impossible to have $a_1b=a_2b$ (you have to do a little casework to cover all the cases where some of the elements might be $1$).
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