0
$\begingroup$

Let $Y$ be a compact, Hausdorff topological space, and $X$ be a locally compact, contractible, Hausdorff space which is homeomorphic to a dense subset of $Y$.

Question A: Is $GL_1(C(Y))\stackrel{\pi}{\longrightarrow} K_1(C(Y))$ surjective?

Question B: If $r:C(Y)\rightarrow C(Y\setminus X)$ denotes the restriction map, then does the following diagram commute? $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} GL_1(C(Y)) & \ra{r} & GL_1(C(Y\setminus X)) \\ \da{\pi} & & \da{\pi} \\ K_1(C(Y)) & \ras{K_1(r)} & K_1(C(Y\setminus X)) \\ \end{array} $$

Improve this question
$\endgroup$
9
  • $\begingroup$ If $Y$ is compact, Hausdorff, and contractible then $K_1(C(Y)$ is trivial... $\endgroup$
    – Paul Siegel
    Commented Nov 11, 2015 at 12:06
  • $\begingroup$ Sorry, my mistake. I meant $X$ is contractible. Will change that now. $\endgroup$
    – Tsol Edud
    Commented Nov 11, 2015 at 12:10
  • $\begingroup$ $GL_1(C(Y))$ corresponds to the set of nonvanishing continuous functions on $Y$, and for connected $Y$ this deformation retracts onto $\{1, -1\}$. So the image in $K_1(C(Y))$ is always trivial, unless I'm missing something. $\endgroup$
    – Paul Siegel
    Commented Nov 11, 2015 at 12:44
  • $\begingroup$ Thanks. I am not sure I follow, since $Y$ is not assumed to be connected? $\endgroup$
    – Tsol Edud
    Commented Nov 11, 2015 at 12:53
  • $\begingroup$ Adding additional connected components will not help $\pi$ become surjective. $\endgroup$
    – Paul Siegel
    Commented Nov 11, 2015 at 13:02

1 Answer 1

Reset to default
2
$\begingroup$

Regarding Question A, the $3$-torus provides a counterexamle: its complex $K_1$-group has rank $4$ but its cohomotopy has only rank $3$. (I have learnt about this example from the book of Rordam, Larsen and Laustsen.)

The diagram in Question B clearly commutes because $K_1(r)$ just acts by picking some representative, restricting it and then taking its $K_1$-class.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .