3
$\begingroup$

My question is prompted by 57589

If $X$ is an object in a monoidal category with unit $I$ then $Y$ is a left dual if we have $I\rightarrow Y\otimes X$ and $X\otimes Y\rightarrow I$ which satisfy the well-known zig-zag identities.

My question is: what is the homotopy version of this story? This is supposed to be equivalent to: if $X$, $Y$ and $C$ are complexes such that $Y$ is left dual to $X$ and $C$ is homotopy equivalent to $Y$ then what structure does $C$ have?

This example is evidently not an operad but it is a PROP. So, a subsidiary question is whether there is a theory of quadratic PROPS and Koszul PROPS which includes this example?

Another subsidiary question is whether the Spanier-Whitehead dual of a space is a dual in this sense. Note that the Spanier-Whitehead dual of a space is only defined up to stable homotopy.

$\endgroup$

1 Answer 1

3
$\begingroup$

The $\infty$-categorical version of the story is identical to the one you wrote (and explained in Lurie's Higher Algebra, Section 4.2.5): in any monoidal $\infty$-category there's a notion of left and right duals of objects, defined by data of evaluation and coevaluation maps together with zig-zag identities, and the left dual if it exists is unique (up to a contractible space of choices). There is thus no extra homotopical flavor to duals (in the $\infty$-setting) -- it's equivalent to have an evaluation map that extends to a duality in the homotopy category. So a choice of your homotopy equivalence of $C$ to $Y$ will give rise to duality data for $C$, exhibiting it as THE dual, and allowing you to make this choice of equivalence unique (as usual up to contractible choices).

I can't comment authoritatively on the S-W dual, and I'm sure others will, but I think it will be the dual of the suspension spectrum of $X$ in the monoidal $\infty$-category of spectra, IF $X$ is dualizable! - which is in this case the same as being a compact object, which means a finite colimit of spheres (eg finite CW complex).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .