Products, internal homs and Hedetniemi's Conjecture - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T04:38:30Zhttp://mathoverflow.net/feeds/question/107591http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/107591/products-internal-homs-and-hedetniemis-conjectureProducts, internal homs and Hedetniemi's ConjectureJulian Kuelshammer2012-09-19T16:33:50Z2012-09-19T21:57:49Z
<p>I'm looking for interesting examples and non-examples of the following property:</p>
<blockquote>
<p>Let $\mathcal{C}$ be a category admitting finite products. An object $K$ has the property (*) if:
$$\forall C,D\in \mathcal{C}(\exists C\times D\to K\Rightarrow \exists C\to K\ or\ \exists D\to K).$$</p>
</blockquote>
<p>If the category has internal homs then by some abstract nonsense this can be reformulated as:</p>
<blockquote>
<p>$K$ has the property that $\nexists C\to K$ then $\exists hom(C,K)\to K$.</p>
</blockquote>
<p>Of course the category should be non-additive (otherwise every object $K$ has property (*) since the product is a coproduct).</p>
<p>The question arose from the subject of graph theory:
$K_n$ has property (*) is equivalent to <a href="http://garden.irmacs.sfu.ca/?q=op/hedetniemis_conjecture" rel="nofollow">Hedetniemi's Conjecture</a> (that if the categorical product of two graphs is $n$-colorable then one of the graphs is $n$-colorable)</p>
<p>I'm also interested in answers dealing with the tensor product instead of the product.</p>
http://mathoverflow.net/questions/107591/products-internal-homs-and-hedetniemis-conjecture/107623#107623Answer by Todd Trimble for Products, internal homs and Hedetniemi's ConjectureTodd Trimble2012-09-19T21:57:49Z2012-09-19T21:57:49Z<p>Any category with a zero object (e.g., pointed sets) has the property. </p>
<p>Any category with a terminal object $1$ such that every object $C$ is either initial or has a point $1 \to C$ has the property. For example, the category of sets or of topological spaces. </p>
<p>Most toposes do not satisfy the property: if $C$ and $D$ are subterminal, then $K = C\times D = C \wedge D$ usually produces a counterexample. </p>