Timeline for Possible semantics for categorical co-constness
Current License: CC BY-SA 3.0
10 events
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| May 4, 2011 at 15:37 | history | edited | Finn Lawler | CC BY-SA 3.0 |
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| Apr 20, 2011 at 13:31 | comment | added | Finn Lawler | In your first example, B is the terminal object of C, and f factors through it. In your second example, fgx=u' but fhx=v', so fg and fh are not equal. The equivalence of the two definitions of constantness is a straightforward application of the usual facts about representable functors and Yoneda's lemma. I suggest you study those more closely, and try to answer for yourself the question I asked above (Mar. 13). | |
| Apr 20, 2011 at 11:19 | comment | added | Jérôme JEAN-CHARLES | @Finn :I found a counter example (C.E.): Let C has two objects A and B, an arrow f from A to B and the two identities arrows then f is nLab constant (vacuously) but NOT Lawvere. Another C.E. is A=x,y ,B={u,v} ,C={u',v'} , T={t} as morphisms add all applications to T (the terminal element). Add morphisms f,g,h with f(u)=u',f(v)=v',g(x)=u ,g(y)=v,h(x)=v ,h(y)=u. Now clearly f(g)=f(h):f is nLab-constant but not Lawvere. More generally if g and h have the same B-fibers it gives a C.E. The idea is nLab means "absorbing" , Lawvere is "one valued". | |
| Apr 20, 2011 at 11:15 | comment | added | Jérôme JEAN-CHARLES | @Finn :I found a counter example (C.E.): Let $C$ has two objects $A$ and $B$, an arrow $f$ from $A$ to $B$ and the two identities arrows then $f$ is nLab constant (vacuously) but NOT Lawvere. Another C.E. is $A={x,y}$ ,B={u,v} ,C={u',v'} , T={t}$ as morphisms add all applications to $T$ (the terminal element). Add morphisms $f,g,h$ with $f(u)=u',f(v)=v'$ , $g(x)=u ,g(y)=v$ , $h(x)=v ,h(y)=u$. Now clearly $f(g)=f(h)$ : $f$ is nLab-constant but not Lawvere. More generally if $g$ and $h$ have the same $B$ fibers it gives a C.E.. The idea is nLab means "absorbing" , Lawvere is "one valued". | |
| Mar 13, 2011 at 16:58 | comment | added | Finn Lawler | Do you understand why it is that if a morphism k is nLab-constant in C then the natural transformation $C(-,k)$ is Lawvere-constant in $[C^{\mathrm{op}}, \mathrm{Set}]$? | |
| Mar 13, 2011 at 1:11 | comment | added | Jérôme JEAN-CHARLES | @Finn : Sorry I still do not see how this show that nlab definition ( on morphisms not functors) implies Lawvere ( assuming only a terminal object) ? | |
| Mar 5, 2011 at 23:16 | history | edited | Finn Lawler | CC BY-SA 2.5 |
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| Mar 5, 2011 at 23:15 | comment | added | Finn Lawler | If each function $f \mapsto kf$ factors through the terminal set, then the natural transformation $k_* \colon C(-,A) \to C(-,B)$ factors through the constant functor at the terminal set. If C has a terminal object T then it represents that functor, so by the full faithfulness of the Yoneda embedding $k = ! \circ b$ for some $b \colon T \to B$, where $! \colon A \to T$. | |
| Mar 5, 2011 at 21:52 | comment | added | Jérôme JEAN-CHARLES | @Finn : The Lawvere definition implies the nlab definition, but i cannot see the reverse implication in a category with a terminal object. | |
| Mar 5, 2011 at 16:55 | history | answered | Finn Lawler | CC BY-SA 2.5 |