Why is the bar construction of a DG algebra a coalgebra?

Question

Let $A$ be a differentially graded augmented algebra. Then $\mathbf{B}A$ can be equipped with the structure of a coalgebra. This is proved in, for example, Loday and Vallette's book on Algebraic Operads.

The $n$-lab page https://ncatlab.org/nlab/show/bar+and+cobar+construction points out that the coalgebra structure can be understood as showing all the possible ways to decompose a path of length $n$ into pairs of shorter paths of lengths $(p,q)$ with $p+q=n$. This helps but I still don't really have intuition.

Many books define $\mathbf{B}A$ as follows. First, we can construct the free (conilpotent) coalgebra on $\mathbf{B}A$. Now, define the differential using the universal property of the free coalgebra, and check that it happens to satisfy the law $d^2=0$; so we have a DG coalgebra.

This makes sense, of course. However even if you had no idea that $\mathbf{B}A$ was supposed to be a coalgebra, there's still a very natural differential to put on that chain complex, which comes from applying the Moore normalization to the simplicial object constructed using this process. https://ncatlab.org/nlab/show/two-sided+bar+construction

So my question is, if I look at $\mathbf{B}A$ as the Moore normalization of the two-sided bar construction $B(1,A,1)$ (where $A$ acts on $1$ as a module on both the left and the right) why should we expect this to have a coalgebra structure. Or, alternatively, if you prefer to view $\mathbf{B}A$ as a coalgebra a priori on top of which we can put a unique compatible differential, why does this differential agree with the one given by $B(1,A,1)$?

I think I am either looking for geometric intuition for why this should be true, or ideally kind of nice categorical argument by which we can construct the comultiplication and counit on $B(1,A,1)$.

Answer 1

This is the type of question with multiple correct answers, because, as you say, it depends very much on what you think the bar construction "is" initially.

You say you want to think of $\mathbf{B}A$ as $B(1,A,1)$. Well, I don't know what you think $B(X,A,Y)$ "is", but one thing that it "is" is a specific nice resolution of the derived tensor product $X \otimes_A^{\mathbb{L}} Y$.

From this perspective, your question is a version of: is there a good reason for $1 \otimes_A 1$ to be a coalgebra? In other words, can we produce a map $$1\otimes_A 1 \to (1 \otimes_A 1) \otimes (1 \otimes_A 1)?$$ Yes. The left hand side is equivalent to $1 \otimes_A A \otimes_A 1$, whereas the right hand side is equivalent to $1 \otimes_A 1 \otimes_A 1$. The (augmentation) ring homomorphism $\epsilon: A \to 1$ that you start with is in particular an $A$-bimodule homomorphism, so you do indeed have the map $$ 1 \otimes_A \epsilon \otimes_A 1 : 1 \otimes_A A \otimes_A 1 \to 1 \otimes_A 1 \otimes_A 1.$$

Then coassociativity is just that you apply $\epsilon$ in "different spots". Oh, and it requires associativity of $\otimes_A$, which I have already suppressed from the notation.

Actually, because $B(-,A,-)$ is a derived tensor product, the associativity of $\otimes_A$ could a priori require lots of coherence data — it could have been merely $\mathsf{A}_\infty$ rather than associative. The result is that you could have concluded that $B(1,A,1)$ was merely a co$\mathsf{A}_\infty$-coalgebra, rather than a coassociative coalgebra. That it is indeed coassociative is a niceness property of the specific resolution that you chose.