Recall that $A(X)$, the K-theory of a connected, pointed space X, is defined as the K-theory spectrum of the ring spectrum $\Sigma^\infty\_+ \Omega X$ (or via a plethora of alternative definitions). Is it known if the homotopy type of $A(X)$ determines the homotopy type of $X$? If not, what is the best one can hope for?


Of course, since $X$ is connected the space $\Omega X$ with its loop space structure determines the homotopy type of $X$, but I am not sure if this is still true when we take $\Sigma^\infty\_+$, I am worried we get might get $X$ only up to $\Sigma^n \Omega^n $. Then there is the question of if ring spectra of this type can have the same K-theory, perhaps we should assume $X$ simply connected to get a positive answer?



Does Waldhausen K-theory detect homotopy type?
The answer to the question



> 
> Does the homotopy type of $()$ determine the homotopy type of $$?
> 
> 
> 


is No in general. As you say, $A(X)$ is determined by the homotopy type of $\Sigma^\infty \Omega X\_+$ as an associative (or $A\_\infty$) ring spectrum, and this ring spectrum does not uniquely determine $X$, even if $X$ is simply-connected.


For example, suppose $Z$ is a pointed space. Let $T(Z)$ be the free associative $S$-algebra generated by $Z$. I.e.,
$$T(Z)=\Sigma^\infty S^0\vee \Sigma^\infty Z \vee (\Sigma^\infty Z)^{\wedge 2} \vee \cdots .$$


If $Z$ is connected, there is an equivalence of associative ring spectra
$$ T(Z) \simeq \Sigma^\infty \Omega\Sigma Z\_+.$$
The equivalence is a version of the classical James splitting. It is induced by a map of spectra $\Sigma^\infty Z \to \Sigma^\infty \Omega\Sigma Z\_+$, extended to a map of ring spectra $T(Z)\to \Sigma^\infty \Omega\Sigma Z\_+$ using freeness.


It follows that if $X$ and $Y$ are connected spaces such that $\Sigma X$ and $\Sigma Y$ are not equivalent, but $\Sigma^\infty X$ and $\Sigma^\infty Y$ are equivalent, then there is an equivalence $A(\Sigma X)\simeq A(\Sigma Y)$ providing a counterexample.


A couple of comments:


* It is well-known that there exist non-isomorphic groups $G$ and $H$ such that the group rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are isomorphic. There are even examples with finite $G$ and $H$. One may wonder if for some of these examples the spherical group rings $\Sigma^\infty G\_+$ and $\Sigma^\infty H\_+$ are equivalent as associative ring spectra. If yes, then $BG$ and $BH$ would provide another counterexample.
* In general one can have non-equivalent ring spectra that have equivalent $K$-theories. For example, I believe that if $P$ and $Q$ are Morita equivalent in a suitable sense, then $K(P)\simeq K(Q)$. Can there be two spaces $X$ and $Y$ such that $\Sigma^\infty \Omega X\_+$ and $\Sigma^\infty \Omega Y\_+$ are not equivalent as ring spectra, but have equivalent categories of modules (in a strong enough sense to induce equivalence of $K$-theories)? It seems far fetched, but I don't know how to exclude this possibility. **Added later**: [A paper by Roggenkamp and Zimmerman](https://pdf.sciencedirectassets.com/271593/1-s2.0-S0022404900X00230/1-s2.0-002240499500121C/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEIP%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJHMEUCIQDPvFmhYCSH3uV39DkXovveN%2BOyDaQDB6Z%2F1kR3sS000gIgWfYmxwaYC3yWTlyGjTWWxBqI584z8DzOVDzIJUJcaEkqgwQI6%2F%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAEGgwwNTkwMDM1NDY4NjUiDBSq2vFuLcecgiZDLirXAxDdL61ZrE4CZZcDfZvm6EeIjuIa1QUcLQpVO%2FnmGP4LboOLwHrmdToaRfYdRJsZKz1HYhCQ7vIcxrVZs7qfObCO9tCadIM8ROh1oIDyhbLkNRW3C6qwBnGQkOKTVGv9DLNCbszGXpbIDGk3%2FcPbPjo1n61XEGdjxcghedRSvYRAB2RRiGxX3ztHT42JLHZ0OJQPE2Wajhu7lI%2FYZeaLOXrZpKtTEVxUuaYa5CkHYaGJz590AaiAU1ayM%2FfrxjmJ%2FW9Xl2ninYRjDhz6Um7DTQPL1jL4Jr10gmSW6Sq9nWJSIhTKrN7GW49zKpJ%2BoVdD0ADeQB2DoE9Wh%2BgrEU%2BTZUKSRiDTRG8Hb7dy5dQO5Q4WsGqBGqkkrsmhU9u2vA6gVrKmhhkoeGA5PZZcHH32%2FWJJJr8K4IAfoDsqlZ21ycgEbb%2BncWPh9a8bGWLrU9i31vorMCaCvXtd21IEs8E8LX7qGXfZIs1L58ujJf8%2F0PyC8L%2FtxP%2FxB1DbgfxQU%2FJGAMuAdzTZ1mwxuBjgkPOSGmrerMyaOl8PyDWLlQghHOcOjHwRvNApCWmBM6SndIApBMoGaja16zQCUuZj4dvw9R8yyNt8Nt6JcK2eVai65y29JERsXqr3GDD7utuKBjqlAevKM%2Budwrs6ydAxjMs72xYyZttDS0qr4tyI0u7b8Pldqv2v1MqbxlbogF1m7Fj%2FoPhQ0r8QbHfCYP1YHGBCZzMiBzHUIGKWQT2bkVevTkSWhdueB80ZcVAVe5rIWd8%2FBHBu1jRpHOXi2uJQRWO3lFHkHnQvVpoE%2BrMvSS9%2Fut%2B75YFa31jvvM30fN5p68pVEeUP%2F84Bc0gHIiiXiiI8gfEs35g%2BKg%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211001T102935Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYYFQ4EQ6V%2F20211001%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=24c11fc3ebd73feb971192d31deb20e90feac3f06e41b8a1184a48754f9c0a91&hash=44d93e5ecbbfbf4421a06fd1b82235a6df599e452d5f327646c6c2b115f5f87d&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=002240499500121C&tid=spdf-f84bc980-2baf-4830-a450-6a9fdfd04fa3&sid=912d2868500e4540d55b51c-0b6669052cf9gxrqb&type=client) gives an example of two groups $G$ and $H$ for which the rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are not isomorphic, but Morita equivalent. It follows that the Quillen $K$-theory of these rings is isomorphic. One may ask whether the $K$-theory spectra of $\Sigma^\infty G\_+$ and $\Sigma^\infty H\_+$ are equivalent as well.


