Suppose I have a commutative ring $R$. Given an element $(x_1,x_2)\in R^2$ there exists a homomorphism $\mathbb{Z} \to R\otimes R$ taking $1$ to $x_1\otimes x_2$, so there exists a map $f:S^0 \to HR \wedge HR$ giving this on $\pi_0$. There is also a map $g:S^0 \to H\mathbb{Z}$ which is an isomorphism on $\pi_0$. Does there exist a map $H\mathbb{Z} \to HR \wedge HR$ that makes this commute, i.e. that takes $1$ to $x_1\otimes x_2$ after applying $\pi_0$?

I'm guessing that in general the answer is "no," but that there might be some kind of obstruction theory that will tell me that sometimes it is "yes," but I'm not sure what it is.


1 Answer 1


You can consider $HR$ as a module over $H=H\mathbb{Z}$, so for any $u\colon S^0\to HR\wedge HR$ you can consider the composite $$ u' = (H \xrightarrow{1\wedge u} H\wedge HR \wedge HR \xrightarrow{\mu\wedge 1} HR\wedge HR) $$ Then the composite $S^0\xrightarrow{\eta}H\xrightarrow{u'}HR\wedge HR$ is the same as $u$, so $u$ and $u'$ have the same effect on $\pi_0$.

Here I used the multiplication of $H$ with the first factor $HR$, but I could have used the second one instead, and that would probably give a different $u'$. This gives some non-uniqueness, and there is probably plenty more from other sources as well.


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