# Augmented Algebras

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 $\mathbb{K}1_A\oplus \ker\epsilon$?

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).
• 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. Commented Dec 4, 2015 at 20:48