Timeline for Morphisms in K-theory: comparison of two pictures
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2016 at 20:57 | vote | accept | truebaran | ||
| Sep 24, 2016 at 1:08 | comment | added | A K | There was an inaccuracy there, thank you! However, there is a simpler argument. Let $v_j \in A^n$ be vectors of the form $(0, \ldots , 0, 1, 0, \ldots , 0)$ with "1" in the position $j$ and zeroes elsewhere, and $w_j$ be analogous vectors in $B^n$. We have $\alpha(v_j) = w_j$ and $\alpha(e v_j) = f(e) w_j$. $f(e)B^n$ is generated by the vectors $f(e)w_j$, which lie in the image of $\alpha$, so we are done. | |
| Sep 24, 2016 at 0:53 | history | edited | A K | CC BY-SA 3.0 |
Clarified meaning.
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| Sep 24, 2016 at 0:47 | history | edited | A K | CC BY-SA 3.0 |
Fixed an inaccuracy in the proof
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| Sep 23, 2016 at 23:04 | comment | added | truebaran | Thank you for your answer. Of course I should have written $A^n$ and $B^n$ instead of $M_n(A)$ and $M_n(B)$. Let me also ask: you showed that if you pick an element in $eA^n$ the image will lie in $f(e)B^n$. However I don't see why the image should be exactly $f(e)B^n$, it seems to me that you have just repeated the same argument as for the fact that $(a_j)_j \in eA^n$ gives that $(f(a_j))_j \in f(e)B^n$ (arguing by contraposition) | |
| Sep 23, 2016 at 22:10 | history | answered | A K | CC BY-SA 3.0 |