most recent 30 from http://mathoverflow.net 2013-05-22T16:11:26Z http://mathoverflow.net/feeds/question/103141 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103141/what-is-the-functor-tensor-product Mozibur Ullah 2012-07-26T01:16:32Z 2012-07-27T17:56:37Z

<p>I'm familiar with the tensor product of modules, but I've also come across functor tensor product (in emily riehls paper on homotopy limits), what are they, and how are they (if they are) related to traditional tensor products? (Emily shows that they can be defined as a particular coend, but that doesn't really provide any intuition for me). </p>

http://mathoverflow.net/questions/103141/what-is-the-functor-tensor-product/103155#103155 Mike Shulman 2012-07-26T04:48:29Z 2012-07-26T04:48:29Z

<p>The <em>enriched</em> version of the functor tensor product does literally generalize the tensor product of modules. A ring is the same as an $\mathbf{Ab}$-enriched category with one object, a (covariant or contravariant) $\mathbf{Ab}$-enriched functor from a ring to $\mathbf{Ab}$ is a (left or right) module, and the $\mathbf{Ab}$-enriched tensor product of such functors is exactly the classical tensor product of a left and a right module.</p>

http://mathoverflow.net/questions/103141/what-is-the-functor-tensor-product/103227#103227 Martin Brandenburg 2012-07-26T18:17:56Z 2012-07-26T18:17:56Z

<p>A very comprehensive account on the tensor product of functors can be found in Section 2.4 of the paper "The fundamental pro-groupoid of an affine 2-scheme" by Alex Chirvasitu and Theo Johnson-Freyd (<a href="http://arxiv.org/pdf/1105.3104v4.pdf" rel="nofollow">arXiv</a>).</p>

http://mathoverflow.net/questions/103141/what-is-the-functor-tensor-product/103264#103264 Peter May 2012-07-27T02:46:10Z 2012-07-27T17:56:37Z

<p>It is easy to be explicit. Not in full generality, given a (small) closed symmetric monoidal category $\mathcal C$ with coequalizers, a covariant functor $M\colon \mathcal C\to \mathcal C$ and a contravariant functor $N\colon \mathcal C \to \mathcal C$, the tensor product $N\otimes_{\mathcal C} M$ is the coequalizer of the diagram </p> <p><code>\[ \coprod_{(c,d)} N(d) \otimes \mathcal C(c,d) \otimes M(c) \implies \coprod_{e} N(e)\otimes M(e).\]</code></p> <p>Here $c,d,e$ range over the objects of $\mathcal C$ and $\implies$ indicates a pair of arrows; one is given by the evaluation maps $N(d) \otimes \mathcal C(c,d)\longrightarrow N(c)$ of $N$ and the other by the evaluation maps $\mathcal C(c,d)\otimes M(c) \longrightarrow M(d)$ of $M$. The similarity to Mike's special case should be clear. This is of course an example of a coend, but I prefer to use the tensor product notation in this special case to make the intuition clear. </p>