conditions for natural transformations to exist? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:01:51Z http://mathoverflow.net/feeds/question/35731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist conditions for natural transformations to exist? EricForgy 2010-08-16T06:17:01Z 2010-08-16T13:31:55Z <p>Given a category $C$ with two objects and one non-identity morphism </p> <p>$$a\to b$$</p> <p>and another similar category $D$ </p> <p>$$x\to y,$$</p> <p>we can define two functors $F,G:C\to D$ with</p> <p>$$F:a\mapsto x, b\mapsto y$$</p> <p>and</p> <p>$$G:a\mapsto x, b\mapsto x$$</p> <p>with morphisms doing the only thing they possibly can.</p> <p>A natural transformation $\alpha:F\Rightarrow G$ would require a component $\alpha_b:F(b)\to G(b)$, but there is no morphism $y\to x$, so if I understand this correctly, there is no natural transformation from $F$ to $G$.</p> <p>Is that correct? Is there a clear set of criteria required for there to exist a natural transformation?</p> http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist/35761#35761 Answer by Aleks Kissinger for conditions for natural transformations to exist? Aleks Kissinger 2010-08-16T13:31:55Z 2010-08-16T13:31:55Z <p>In general, there are probably no conditions for the existence of natural transformations that are simpler than just using the definition of naturality itself. In a category Z with a zero object, the zero natural transformation between functors $F,G : C \rightarrow Z$ always exists, but this is quite a degenerate example.</p> <p>Often, you can arrive at an intuition as to whether a natural transformation exists if you consider what such a thing would mean in the specific categories involved. For instance, functors from $(\bullet \rightrightarrows \bullet)$ into Set are just graphs. A natural transformation $G\Rightarrow H$ then exists precisely when there is a graph homomorphism from $G$ to $H$.</p>