Let $(E,F,\alpha)$ represent an element of the second group. Wlog $\alpha:E\to F$ is defined globally even though it's only an iso outside $K$. Choose global compactly supported sections $s_1,\dots ,s_n$ of the dual of $E$ such that on $K$ they span the bundle; for large enough $n$ this is possible. Let $E'$ be the trivial rank $n$ bundle and interpret these sections as an injective map $s:E\to E'$. We now have a bundle map $(\alpha,s):E\to F\times E'$ which is injective everywhere. Split off the resulting subbundle of $F\times E'$ and call the other part $F'$. This leads to an isomorphism $E\times F'\cong F\times E'$for some bundle $F'$. . Replace $(E,F,\alpha)$ by $(E\times E',F\times E',\alpha\times 1)$. Near infinity, where $\alpha\times 1$ is an iso, the composed iso $E\times E'\to F\times E'\to E\times F'$ looks likes $(1, 0)$ on the $E$ part and therefore looks like $(?,\alpha')$ on the $E'$ part where $\alpha'$ is an iso. Make a little adjustment so that it looks like $(0,\alpha')$ on the $E'$ part. This iso $\alpha':E'\to F'$ near infinity yields $(E',F',\alpha)$ (E',F',\alpha')$ equivalent to $(E,F,\alpha)$, but with the bundles trivial near infinity. (in In fact , $E'$ is trivial globally).globally.)
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Let $(E,F,\alpha)$ represent an element of the second group. Wlog $\alpha:E\to F$ is defined globally even though it's only an iso outside $K$. Choose global compactly supported sections $s_1,\dots ,s_n$ of the dual of $E$ such that on $K$ they span the bundle; for large enough $n$ this is possible. Let $E'$ be the trivial rank $n$ bundle and interpret these sections as an injective map $s:E\to E'$. We now have a bundle map $(\alpha,s):E\to F\times E'$ which is injective everywhere. Split off the resulting subbundle of $F\times E'$ and call the other part $F'$. This leads to an isomorphism $E\times F'\cong F\times E'$ for some bundle $F'$. Replace $(E,F,\alpha)$ by $(E\times E',F\times E',\alpha\times 1)$. Near infinity, where $\alpha\times 1$ is an iso, the composed iso $E\times E'\to F\times E'\to E\times F'$ looks likes $(1, 0)$ on the $E$ part and therefore looks like $(?,\alpha')$ on the $E'$ part where $\alpha'$ is an iso. Make a little adjustment so that it looks like $(0,\alpha')$ on the $E'$ part. This iso $\alpha':E'\to F'$ near infinity yields $(E',F',\alpha)$ equivalent to $(E,F,\alpha)$, but with the bundles trivial near infinity (in fact, $E'$ is trivial globally). |
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