Suppose $c: X \to X\vee X$ is a co-$H$ structure on a based CW complex $X$.

** Question:** Under what circumstances can one find left (right) homotopy inverses for $c$?

* Remarks: * If $X$ is $1$-connected, then the answer is yes (for a proof, see below).
But I don't know the answer in general. A good test case would be to take an exotic comultiplication on the circle and see what happens there.

A reduced suspension of a based space, with its tautological comultiplication (the pinch map), has homotopy inverses. These arise from the reflection map on the circle coordinate.

** Definitions: ** a * co-*$H$ structure on $X$ is a map $c$ as above such that
the composite
$$
X \quad \overset{c}\to \quad X \vee X \quad \overset{\text{include}}\longrightarrow \quad X \times X
$$
is homotopic to the diagonal. A * left homotopy inverse * for $c$ is a map
$\ell: X \to X$ such that the composite
$$
X \quad \overset{c} \to \quad X\vee X\quad \overset{\ell \vee \text{id}}\longrightarrow \quad X\vee X
\quad \overset{\text{fold}}\longrightarrow \quad X
$$
is homotopic to a constant map. (Right homotopy inverses are defined similarly.)

* proof when $X$ is $1$-connected: * Form the * coshearing map *
$$
\check S : X \vee X \to X \vee X
$$
which on the first summand of the domain is given by $c$ and on the second by the second summand inclusion. It is straightforward to check that $\check S$ is a homology isomorphism,
so by the Whitehead theorem, there's a map $T: X\vee X \to X\vee X$ such that
$\check S\circ T$ and $T\circ \check S$ are each homotopic to the identity map.

Then the composite $$ X \quad \overset{i_1}\to \quad X \vee X \quad \overset{T}\to \quad X\vee X \quad \overset{p_2}\to \quad X $$ is a left homotopy inverse for $c$, where $i_1$ is the first summand inclusion and $p_2$ is the projection onto the second summand.