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I'm a newcomer to operads so apologies if this is a naive question.

The standard picture of an operad is of a collection of $n$-ary operations, thought of as objects with $n$ upward-pointing legs (inputs) and one downward-pointing leg (output), which can be composed in a sensible way. "Algebras over operads" then give rise to all sorts of algebras in the usual sense (associative, $A_\infty$ etc).

My understanding is that a PROP is basically the same thing but with operations allowed to have more than one output. Algebras over PROPs can also accommodate coproduct-like operations, and hence also include useful things like coalgebras and Hopf algebras.

Historically though, operads came later than PROPs. So why was it useful to single out the notion of an operad, as opposed to a general PROP?

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  • $\begingroup$ I would guess the examples of interest only required the operadic structure. PROPs are still useful, I guess now people say PROperads are the right notion. $\endgroup$
    – Sean Tilson
    Commented May 12, 2016 at 10:25
  • $\begingroup$ Personally, I would not say operads are "better", because they're less general. Any algebra over an operad is an algebra over a PROP. That said, the generality of PROPs have made them hard to work with in some ways. Some examples of difficulties with abstract homotopy theory (model categories) and PROPs, that are not present for operads, can be found in work of Benoit Fresse. But I think your question is more general than abstract homotopy theory, so I'll leave it to others to answer. $\endgroup$
    – David White
    Commented May 12, 2016 at 13:06

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I was friends with Frank Adams and Saunders Mac Lane, who invented PROPs in one of the world's most extensive unpublished collaborations. Saunders once showed me a box full of their correspondence. One reason they never published is that they lacked a way of showing the PROPs they were interested in acted on the things they wanted to have actions. Operads are of course equivalent to a special kind of PROP, and the specialization made it very much easier to find operad actions. The connection with monads was intrinsic to the definition of operads (I convinced Mac Lane to switch from "triples'' to "monads" in Categories for the Working Mathematician in large part in order to make the names operad and monad to mesh) and operads allowed tons of explicit computations (in the homology of iterated loop spaces in particular) that would not have not come naturally if at all from PROPs. Of course, there are interesting examples of PROPs of the more general sort, ones that do not come from an operad, but they are irrelevant to the original work with iterated loop spaces.

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  • $\begingroup$ Straight from the inventor himself! Thanks :) $\endgroup$
    – user81684
    Commented May 13, 2016 at 7:59
  • $\begingroup$ Is it obvious that the computations for PROPs are not feasible? Doesn't the space of embeddings of k disks in n disks admit a nice decomposition? What mistake(s) am I making? $\endgroup$
    – Sean Tilson
    Commented May 13, 2016 at 19:29
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    $\begingroup$ If you can compute something, go ahead! I was thinking for example of the calculations described here: mathoverflow.net/questions/180637/… $\endgroup$
    – Peter May
    Commented May 13, 2016 at 19:39
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One thing you can do with an operad that you cannot do with a prop is write down a monad such that algebras over the monad correspond to algebras over the operad. For example, Hopf algebras have a prop but not an operad, and don't even have a monad: I believe the forgetful functor from Hopf algebras to vector spaces doesn't have a left adjoint, so cannot be monadic. This means that you can't apply useful results like monadic resolutions, etc. to algebras over props.

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  • $\begingroup$ Right, Qiaochu. One point of the definition of operads is to deliberately avoid operations with repeated variables. $\endgroup$
    – Peter May
    Commented May 13, 2016 at 19:41
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    $\begingroup$ Qiaochu's point would apply equally well to bialgebras, where the point about repeated varaiables is moot. It's a different point. $\endgroup$
    – Todd Trimble
    Commented Jun 11, 2016 at 1:27

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