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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 $(E',F',\alpha)$ (E',F',\alpha')$equivalent to$(E,F,\alpha)$, but with the bundles trivial near infinity. (in In fact ,$E'$is trivial globally).globally.) 1 Let$(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).