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My question is on the example below, taken from page 4 of http://www.math.wisc.edu/~andreic/publications/lnPoland.pdf.

I'm not familiar enough with this stuff yet to understand why the quasi-isomorphism below does not have an inverse.

I'm also caught up on the second example since I don't understand what the map $(x, y)$ means. Is that projection from the first part of the direct sum onto $\mathbb{C}[x, y]$?

I would be appreciative if anyone here could work out the details on these examples.

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  • $\begingroup$ There are no nonzero maps from $Z/2$ to $Z$, for the first question. $\endgroup$ – Kevin Carlson Apr 22 at 16:55
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    $\begingroup$ The obvious one is $(f,g)$ goes to $xf+yg$, but then they are quasi-isomorphic (from the second complex to the first: $f \to (yf,-xg)$, $z\to z$). $\endgroup$ – A.G Apr 22 at 20:06

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