most recent 30 from http://mathoverflow.net 2013-05-19T00:33:10Z http://mathoverflow.net/feeds/question/120120 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map Simone Virili 2013-01-28T15:51:27Z 2013-05-07T00:22:00Z

<p>In [M. Gromov, Endomorphisms of symbolic algebraic varieties, J. Eur. Math. Soc. (JEMS) 1 (1999), 109–197], Gromov introduces the notion of pre-injective map. Recasting this notion in the setting of Abelian groups we get the following:</p> <p>Definition. Given a finite discrete Abelian group $G$, an (infinite) index set $X$ and taking the product $G^X$ endowed with the product topology, we say that a continuous endomorphism $\phi:G^X\to G^X$ is <em>pre-injective</em> if the restriction of $\phi$ to $G^{(X)}$ (the subgroup of elements with finite support) is injective. </p> <p>The Pontryagin-Van Kampen dual of $\phi$ is just an endomorphism $\widehat\phi$ of the discrete group $G^{(X)}$. Can we say that $\phi$ is pre-injective just looking at $\widehat\phi$? In other words, what is the algebraic property of $\widehat\phi$ which corresponds to the pre-injectivity of $\phi$? </p>

http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map/120146#120146 Phu Chung 2013-01-28T20:23:38Z 2013-01-28T20:32:32Z

<p>Well, in the case X is also a group, your question somehow relates with cellular automata, i.e your map <code>$\phi :G^X\rightarrow G^X$</code> is continuous and X-equivariant. When X is amenable,<code>$\phi$</code> is pre-injective if and only if <code>$\phi$</code> is surjective. Thus, your dual <code>$\widehat{\phi}$</code> will be injective in this case. It is not true anymore when <code>$X=F_2$</code>, the free group with 2 generators, see section 5.11 of Ceccherini-Silberstein, Coornaert's book "Cellular Automata and Groups" for an example of pre-injective but not surjective cellular automaton over <code>$F_2$</code>.</p>