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If $f_n$ is a sequence of smooth orientation preserning mappings of degree one between open annuli $A(1,r):=\{x: 1< |x| < r \}$ and $A(1,r_n)$, $r>1$ and $r_n>1$, on the Euclidean space $\mathbf{R}^d$, converging (in compacts of $A(1,r)$ together with the derivatives) to a smooth mapping $f$. What can be said about the degree of $f$?

Of course all the mappings $f_n$ are sirjective and for einstance map the inner (outer) boundary onto inner (outer) boundary

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  • $\begingroup$ What does your notation $A(1,r)$ mean? Do you mean the set of points in $R^d$ of norm $\ge 1$ and $\le r$? $\endgroup$
    – Lee Mosher
    Commented May 21, 2012 at 12:51
  • $\begingroup$ $A(r,1)$ is the set of points of norm $>1 $ and $<r$. $\endgroup$
    – Marijan
    Commented May 21, 2012 at 12:54
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    $\begingroup$ Next question: without further restrictions, the degree is not well-defined. For instance, take $r=2$, $r_n=3$ for all $n$, and $f_n : A(1,2) \to A(1,3)$ to be the inclusion map, so $f_n$ converges to the inclusion map. Over all points of $A(1,3)$ of norm $<2$ the degree is $1$, and over all points of norm between $2$ and $3$ the degree is zero, so the degree of the limit is not well-defined. Do you have any further restrictions in mind to avoid examples like this? $\endgroup$
    – Lee Mosher
    Commented May 21, 2012 at 13:14
  • $\begingroup$ Of course all the mappings $f_n$ are sirjective and for einstance map the inner (outer) boundary onto inner (outer) boundary. $\endgroup$
    – Marijan
    Commented May 21, 2012 at 13:33
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    $\begingroup$ Degree of the limit is 1: Just stretch all target annuli to be the same. The new maps still converge and are eventually homotopic to limit rel. boundary. $\endgroup$
    – Misha
    Commented May 21, 2012 at 14:03

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