Not a good title.

Suppose we have two dg symmetric Koszul operads, say $O_1$ and $O_2$. Then their (homotopy) algebras over a dg vector space $V$ are (equivalent to) twisting morphisms

$$f\in TW(O^*_{i}, End_V)$$

where $O^*$ is the Koszul dual dg-cooperad. (I don't even know the \TeX symbol for the anti-shriek) and $TW(O^*_{i}, End_V)$ is the solution set of the Maurer Cartan equation with respect to the dg Lie algebra structure on $Hom_{S-mod}(O^*_{i}, End_V)$.

Now any (infinity) morphism of dg Lie algebras

$$G: Hom_{S-mod}(O^*_1, End_V) \to Hom_{S-mod}(O^*_2, End_V)$$

maps solutions of the Maurer Cartan equation into solutions of the Maurer Cartan equation, therefore transforming (homotopy) $O_1$-algebras into (homotopy) $O_2$-algebras.

One way to find some of these dg Lie algebra maps, is via cooperad morphisms: Suppose we have an (infinity) morphism of dg-cooperads

$$F: O^*_2 \to O^*_1.$$

Then $F$ induces an (infinity) morphism of dg-Lie algebras

$$G_F:= F^*: Hom_{S-mod}(O^*_1, End_V) \to Hom_{S-mod}(O^*_2, End_V)$$

via pull-back.

Now my question is:

Is any (infinity) Lie algebra morphism $G$ induced by some (infinity) cooperad morphism $F$? I mean is there a 1:1 correspondence? Or are there (infinity) dg-Lie algebra maps $G$, that do not come from some (infinity) dg-cooperad map $F$?

• Do you care about naturality in $V$? If not, this looks false. Let $O_1=O_2$ be free on a generator in arity 1, then $O_1^*=O_2^*$ has 1 dimensional coaugmentation cokernel. Both $L_\infty$ algebra structures are $End_V(1)$ (with a degree shift) and just the differential from $End_V$ as $L_\infty$ structure. An $L_\infty$ morphism is a collection of chain maps $End_V(1)^{\otimes n}\to End_V(1)$ of the appropriate degrees, with no further conditions. I can't tell what the formula for $F^*$ is but if I pretend I know what you mean, I don't see how you could possibly hit all such collections. – Gabriel C. Drummond-Cole Oct 1 '16 at 15:02
• I have no clue about your question, but if you want to type an anti-shriek, you can just copy/paste the character for it: ¡ (note that in actual LaTeX it wouldn't work as well, but with mathjax is works fine), e.g. $P^¡$. – Najib Idrissi Oct 3 '16 at 13:08