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I am having problems with this question in Awodey's Category Theory book p.248:

  1. Show that every monoid M admits a surjection from a free monoid $F(X) → M$, by considering the counit of the free $\dashv$ forgetful adjunction.

Let $f : F(X)→M$ be a monoid homomorphism and $m \in M$. We want to show that there is $x \in F(X)$ such that $f(F(x))=m$. But now I'm kind of stuck.

I know I should use the the UMP of the counit $\epsilon : F \circ U \to 1_\mathbf{D}$ somehow, which says that, for $C \in \mathbf{C}$ and $D \in \mathbf{D}$, each $f : F(C) \to D$ determines a function $g : U(D) \to C$ uniquely up to isomorphism such that $f = F(g) \circ \epsilon_D$.

Any help would be highly appreciated!

Thanks!

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  • $\begingroup$ Well, actually, the question says "surjection", so why don't you prove that it's surjective? It will then follow that you have an epimorphism. $\endgroup$ – Zhen Lin Jan 25 '16 at 12:47
  • $\begingroup$ @ZhenLin But isn't "surjection" here ambiguous? For instance, surjective on objects or morphisms (or both)? If it is the former, then it seems to be trivial, since each monoid only has one object. Otherwise, in the case of it being surjective on arrows, I am not sure how could I proceed. I'm editing the question to make this point clear. Thanks! $\endgroup$ – StudentType Jan 25 '16 at 13:47
  • $\begingroup$ For this exercise, I do not think it is helpful to think of monoids as being special categories. If you insist, "surjective homomorphism" here means "full functor". $\endgroup$ – Zhen Lin Jan 25 '16 at 13:57
  • $\begingroup$ @ZhenLin Ah, I understand what you mean. But still I'm stuck to prove that $f$ is surjective. Could you give me some hint or something? : ) $\endgroup$ – StudentType Jan 25 '16 at 14:22
  • $\begingroup$ The monoid homomorphism $F (X) \to M$ you should be considering is not arbitrary. The hint says to think about the adjunction counit. Well, it is a monoid homomorphism – did you try showing it is surjective? $\endgroup$ – Zhen Lin Jan 25 '16 at 14:25
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Let $f : F(X)→M$ be a monoid homomorphism and $m \in M$. We want to show that there is $x \in F(X)$ such that $f(F(x))=m$. But now I'm kind of stuck.

This looks like you're trying to show that every monoid homomorphism from a free monoid is surjective, but you only have to show that there exists a surjection from a free monoid.

You're told to use the counit of the adjunction, which is a natural transformation $FU\rightarrow id$. So its component at $M$ is a homomorphism $F(U(M))\rightarrow M$. Try to show that this is surjection.

(Hint: Find a right inverse to the underlying function $U(F(U(M)))\rightarrow U(M)$.)

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  • $\begingroup$ So is $\epsilon_M : FU(M) \to M$ surjective, because for each $m \in M$, $\epsilon_M (m)=m$? I don't even know if I can assume this. I am kind of lost in my thoughts.. $\endgroup$ – StudentType Jan 25 '16 at 14:37
  • $\begingroup$ Yeah, that's the right answer. But what exactly do you mean by "$m$" on the LHS of $\epsilon_M(m)=m$? When you think about it you'll find that you mean the image of $m$ under the function $M\rightarrow FU(M)$ (or to be more precise $U(M)\rightarrow UFU(M)$ since we mean a function between the underlying sets) given by the unit at $U(M)$. So another way of saying this is that the composition $U(M)\rightarrow UFU(M)\rightarrow U(M)$ is the identity, which you'll find is one of the "triangle equations" for an adjunction. $\endgroup$ – Oscar Cunningham Jan 25 '16 at 14:48
  • $\begingroup$ Mmm... I think I know what you mean! Then I take for granted that since $\eta_{U(M)} : U(M) \to UF(U(M))$ is a function on the carrier of $M$ and $FU(M)$, it uniquely determines a function $M \to F(U(M))$, right? But can I use it to define $f : FU(M) \to M$? Shouldn't we show that it is injective first (or we are just implicitly assuming choice)? $\endgroup$ – StudentType Jan 25 '16 at 15:33
  • $\begingroup$ You know that it's injective because the triangle equation tells you that $U(\epsilon_M)\circ\eta_{U(M)}:U(M)\rightarrow U(M)$ is the identity, and so $\eta_{U(M)}$ is split mono, hence injective. $\endgroup$ – Oscar Cunningham Jan 25 '16 at 15:39
  • $\begingroup$ But more to the point is that the same equation tells you that $U(\epsilon_M)$ is split epi, hence surjective, which is what we are trying to prove! $\endgroup$ – Oscar Cunningham Jan 25 '16 at 15:45

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