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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.

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For the circle, the map of degree $-1$ will always be an inverse. More generally, if $\ell: X\to X$ acts like $-1$ on (reduced) homology, and if the only map $X\to X$ that acts like $0$ on homology is the constant map, then $\ell$ must be an inverse for any co-H-space structure on $X$. – Tom Goodwillie Jan 7 '11 at 15:40
I think Tom's argument shows that any co-H-space structure on a suspension is a co-H-group. On the other hand there are examples of co-H-groups on spaces which aren't suspensions. – Mark Grant Jan 7 '11 at 16:27
John is using "co-H-space"in a weak sense, no associativity requirement. So his exotic structures on the circle do not even make it a comonoid. By the way, I don't know whether inverses are ever worth much without associativity. – Tom Goodwillie Jan 7 '11 at 17:02
I guess my "test case" was too simple minded. – John Klein Jan 7 '11 at 19:14
How about a finite wedge of circles? – John Klein Jan 7 '11 at 20:33

The argument you give for the simply-connected case generalizes to nilpotent connected co-H spaces. This is essentially because nilpotent spaces are $H_\ast(-;\mathbb{Z})$-local. See

Hilton, Peter; Mislin, Guido; Roitberg, Joseph On co-H-spaces. Comment. Math. Helv. 53 (1978), no. 1, 1–14,

in particular Theorem 2.3 and the remarks just after. However, in the same paper they show that the only non-simply-connected nilpotent connected co-H-space is $S^1$, and remark that "it is easy to see that any co-H-structure on $S^1$ admits a 2-sided co-inverse".

I think that if $X$ and $Y$ are co-H-spaces which admit left (right) co-inverses $\ell_X\colon\thinspace X\to X$ and $\ell_Y\colon\thinspace Y\to Y$, then $X\vee Y$ with the co-H-structure $$ X\vee Y \stackrel{c_X\vee c_Y}{\longrightarrow} (X\vee X)\vee(Y\vee Y)\cong (X\vee Y)\vee(X\vee Y) $$ has as a left (right) co-inverse the map $\ell_X\vee\ell_Y$. However this does not seem to be enough to answer your question about a finite wedge of circles, since a given co-H-structure may not split as a wedge in this way.

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