Morphism of conilpotent coalgebras

Question

I have a stupid question about morphisms of two conilpotent coalgebras $\phi:X\to Y$. Is it a morphism of coaugmented coalgebras such that $\phi(F_n(X))\subseteq F_n(Y)$? Here $F_n$ denotes the coradical filtration. This plays a role in the definition of cofreeness. I am a bit lost with the literature.

Answer 1

A coaugmented coalgebra $X$ over a field $k$ is a (nonzero) coalgebra endowed with a morphism of coalgebras $k\to X$. The canonical filtration $F$ on a coaugmented coalgebra $X$ is defined by the rule that, for every $n\ge0$, the subspace $F_nX\subset X$ consists of all elements $x$ such that the image of $x$ in $X/k$ is annihilated by the iterated comultiplication map $X/k\to (X/k)^{\otimes n+1}$.

A coaugmented coalgebra $X$ is said to be conilpotent if its canonical filtration is exhaustive, that is, $X=\bigcup_{n\ge0}F_nX$. In this case, the canonical filtration on $X$ coincides with what is called the coradical filtration of a coalgebra $X$. (The latter is defined similarly except that one uses the quotient of $X$ by its maximal cosemisimple subcoalgebra instead of the quotient by the image of the coaugmentation map. For a conilpotent coalgebra, the maximal cosemisimple subcoalgebra coincides with the image of the coaugmentation map.)

Any morphism of coaugmented coalgebras $\phi\colon X\to Y$ preserves the canonical filtrations, i.e., one has $\phi(F_nX)\subset F_nY$ for any morphism of coaugmented coalgebras $\phi$. This follows immediately from the definition of the canonical filtration. Moreover, one can see that the coaugmentation of a conilpotent coalgebra is unique; hence any morphism of conilpotent coalgebras preserves the coaugmenations.

Accordingly, you can define a morphism of conilpotent coalgebras $\phi\colon X\to Y$ as an arbitrary morphism of coalgebras $X\to Y$, or if you prefer, as a morphism of coaugmented coalgebras $X\to Y$. The condition that $\phi(F_nX)\subset F_nY$ then follows automatically.