First: upper-star, then: lower-star, finally: lower-shriek - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:44:49Z http://mathoverflow.net/feeds/question/35531 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35531/first-upper-star-then-lower-star-finally-lower-shriek First: upper-star, then: lower-star, finally: lower-shriek David Spivak 2010-08-13T19:37:09Z 2010-09-12T00:25:30Z <p>For a category $\mathcal{C}$, let $\mathcal{C}-Set$ denote the category of functors $\mathcal{C}\to{\bf Set}$. Recall that given a functor $F\colon\mathcal{B}\to\mathcal{C}$, the composition with $F$" functor is denoted $F^*\colon\mathcal{C}-Set\to\mathcal{B}-Set.$ It has a left and a right adjoint, $F_!$ and $F_*$. I call these functors the pullback, the left pushforward, and the right pushforward.</p> <p>Let $p\colon A\to B$ be a function of sets, thought of as a functor $P\colon[1]\to{\bf Set}$, where $[1]$ is the "free-arrow category," $[1]="\bullet\to\bullet$." Suppose one wants to find the image of $p$, but he or she can only use pull-backs, left pushforwards, and right pushforwards to manufacture it. In other words, suppose one wants to find a zigzag of functors $[1]=:C_0\leftarrow C_1\rightarrow C_2\leftarrow C_3\rightarrow\cdots\rightarrow C_n=[0]$ such that if we perform a pullback along all leftward functors and either a left pushforward or a right pushforward along rightward functors, then the end result will be the image set $im(p)$ of $p$ (considered as a functor $[0]\to{\bf Set}$).</p> <p>This can be done. To do it, I used a sequence of the form $$[1]\leftarrow C_1\rightarrow C_2\rightarrow [0].$$ If the functors are denoted (left to right) by $F,G,$ and $H$, I found that $H_! \circ G_*\circ F^\ast (P)=im(P)$.</p> <p>I'm not going to bore you with the details of $C_1, C_2$ and $F,G,H$. </p> <p>Here's the question. I've seen things like $H_! \circ G_*\circ F^\ast$ before in the context of polynomial functors. Unfortunately, I don't know enough about them to know if there's a connection. Is there? </p> <p>I also don't know if I can get the whole epi-mono factorization somehow. I haven't worked that long at it, but suppose I want not to end up with the set $im(p)$ but instead the maps $A\to im(f)\to B$. Can I achieve that by use of pullbacks and pushforwards as above (with $C_n=[2]$ now)? Is there any rhyme or reason to such constructions?</p> <p>Thanks. </p> http://mathoverflow.net/questions/35531/first-upper-star-then-lower-star-finally-lower-shriek/38433#38433 Answer by Todd Trimble for First: upper-star, then: lower-star, finally: lower-shriek Todd Trimble 2010-09-12T00:25:30Z 2010-09-12T00:25:30Z <p>First of all: yes, there's certainly a connection. See <a href="http://ncatlab.org/nlab/show/polynomial+functor" rel="nofollow">http://ncatlab.org/nlab/show/polynomial+functor</a>. If the base category is $Set$, the composite </p> <p>$$Set/W \stackrel{f^\ast}{\to} Set/X \stackrel{g_\ast}{\to} Set/Y \stackrel{h_!}{\to} Set/Z$$ </p> <p>first takes a $W$-indexed set $S_w$ to an $X$-indexed set $T_x = S_{f(x)}$, then takes this to the $Y$-indexed set $U_y = \prod_{x: g(x) = y} T_x$, then takes this to the $Z$-indexed set $V_z = \sum_{y: h(y) = z} U_y$. Putting this together, the composite is a family of polynomials, each a sum of monomial terms </p> <p>$$P(\ldots, S_w, \ldots) = (z \mapsto \sum_{y \in h^{-1}(z)} \prod_{x \in g^{-1}(y)} S_{f(x)})$$</p> <p>I'll give a quick example. Suppose we want to express the free monoid functor </p> <p>$$F(S) = \sum_{n \geq 0} S^n$$ </p> <p>in this form. Then we take $W = 1$, $X = \mathbb{N} \times \mathbb{N}$, $Y = \mathbb{N}$, $Z = 1$. There's only one choice for $f$ and $h$, and $g$ is rigged so that the fiber over $n \in \mathbb{N}$ is an $n$-element set: $g(m, n) = m + n + 1$. One can easily check this works. </p> <p>As for the other question: it would have been nice if you had "bored" us! Because I don't see how to reconstruct what you did. What I have to get the image is a zig-zag of length 4 </p> <p>$$Set^{[1]} \stackrel{F^\ast}{\to} Set^{C_1} \stackrel{G_\ast}{\to} Set^{C_2} \stackrel{H^\ast}{\to} Set^{C_3} \stackrel{J_!}{\to} Set^{[0]}$$ </p> <p>where $C_1$ is the generic cospan $a \to c \leftarrow b$, $C_2$ is the generic commutative square, $C_3$ is the generic span $a \leftarrow d \to b$, and then $G$ and $H$ are the evident inclusion functors, and $F$ takes each arrow of the generic span to the arrow of $[1]$. Then $F^\ast$ takes $p: A \to B$ to the cospan consisting of two copies of $p$; hitting this with $G_\ast$ takes this cospan to the pullback square (pulling back $p$ against itself); hitting this with $H^*$ restricts the pullback square to the span consisting of the pullback projections; finally, hitting this with $J_!$ takes this span to its colimit = pushout, which is the same as the coequalizer of the pullback projections (because they have a common right inverse). (Based on his comment, I'm guessing that some guy on the street was doing more or less the same thing.) </p> <p>Could you tell us what you had in mind? </p>