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Let $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k})$ be the category of monoids in the monoidal category of algebras $\textbf{Alg}$, with product $\otimes_{\mathbb{k}}$ and unit element $\mathbb{k}$. I want to show that $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k}) = \textbf{CAlg}$, but the thing I obtained is $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k}) = \textbf{Alg}$.

This is because $X \in \textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k})$ is an algebra with an associative multiplication $X \otimes X \rightarrow X$ and a unit $u: \mathbb{k} \rightarrow X$. So, the multiplication is only associative, not commutative.

How can it turn out to be commutative in order to obtain $X \in \textbf{CAlg}$?

Phi_24
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1 Answers1

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The multiplication $\mu : X \otimes X \rightarrow X$ is also an algebra homomorphism (since it lives in $\mathbf{Alg}$), not just a linear map as in the usual case, so satisfies the relation $\mu (x*x' \otimes y*y' ) = \mu (x \otimes y) * \mu (x' \otimes y')$ where $*$ denotes the algebra product on $X$. This is not true in general if $\mu = *$ but only true if the algebra is commutative. Just set $x = y' = 1$ so we get $x'*y = y*x'$.

The tricky part is proving that $\mu = *$ which you can do with the Eckmann Hilton argument.

Noel Lundström
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