3
$\begingroup$

Recently I started to study Operads. My reference is Algebraic Operads of Jean-Louis Loday and Bruno Vallette. In this book they define augmented algebra of the following form: an $\mathbb{K}$-algebra $A$ is augmented when there is a morphism of algebras $\epsilon: A\rightarrow \mathbb{K}$ called augmentation map.

My problem starts when they claim that if $A$ is augmented then $A$ is canonically isomorphic, as vector space, to $\mathbb{K}1_A\oplus \ker\epsilon$.

Well, I know that the Splitting Lemma gives us an isomorphism between $A$ and $\mathbb{K}1_A\oplus \ker\epsilon$, since that $\epsilon$ is surjective and all exact sequence of vector spaces splits. But the proof that I know of this result depends on the basis of the vectorial spaces and for me it does not provide a canonical isomorphism.

So the book is correct? If is, how do I proof that there is a canonical isomorphism between A and $\mathbb{K}1_A\oplus \ker\epsilon$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Define $\phi \colon A \to \mathbf K1_A \oplus \ker \epsilon$ by $$ \phi(a) = \bigl(\epsilon(a)1_A, a - \epsilon(a)1_A\bigr) $$ then $\phi$ is obviously linear and one-to-one. And if $(r1_A, a) \in \mathbf K 1_A \oplus \ker \epsilon$ is given, then $$ \epsilon(r1_A + a) = r + \epsilon(a) = r $$ hence $$ \phi(r1_A + a) = (r1_A, a) $$ so $\phi$ is onto. Hence, $\phi$ is an isomorphism (and no choices involved in its construction).

$\endgroup$
1
  • $\begingroup$ I want to point out, there was a choice of sorts: the unit $\eta : \mathbb{K} \to A$ sending $1_{\mathbb{K}}$ to $1_A$. It "comes with" $A$ so there isn't really a choice to make and it's why the splitting is canonical when defined on the category of unital augmented algebras, but technically the splitting does depend on it. $\endgroup$ Commented Dec 4, 2015 at 20:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .