1
$\begingroup$

Let $f:X\to Y$ be a (surjective if it helps) map of simplicial commutative monoids such that the induced map on homotopy groups $\pi_n(X,1)\to \pi_n(Y,1)$ is an isomorphism for all $n\geq 0$. Let $x\in X$. Is $\pi_n(X,x)\to \pi_n(X,f(x))$ necessarily an isomorphism? (I don't even know if homotopy groups with basepoint in different connected components of a simplicial commutative monoid can be non-isomorphic).

$\endgroup$
  • 2
    $\begingroup$ $X=S^1\cup\{0\}$ is a simplicial commutative monoid under complex multiplication, right? Yet components have different homotopy groups. Also, to be clear: $f$ is assumed to be a monoid homomorphism? Or just continuous? $\endgroup$ – freakish Feb 8 at 15:39
  • $\begingroup$ $f$ is a homomorphism of simplicial monoids, not just of simplicial sets. $\endgroup$ – A.G Feb 8 at 17:29
  • 2
    $\begingroup$ Anyway the direct counterexample is $M$ is any simplicial connected monoid with nontrivial fundamental group (e.g. $M=S^1$) and you artificially add new identity $X=M\sqcup\{*\}$ and put $f:X\to \{*\}$ the constant map. It induces isomorphisms on $x=*$ but not on $x\neq *$. $\endgroup$ – freakish Feb 8 at 17:31
  • $\begingroup$ @freakish: That doesn't quite work since it won't induce an isomorphism on $\pi_0$. $\endgroup$ – Eric Wofsey Feb 8 at 17:32
  • $\begingroup$ Thank you anyway for the first comment. $\endgroup$ – A.G Feb 8 at 17:37
2
$\begingroup$

No. For instance, let $A$ be any connected simplicial commutative semigroup, and let $X=A\sqcup\{*\}$ with the monoid structure such that $*$ is the identity. Let $B$ be another connected simplicial commutative semigroup and $Y=B\sqcup\{*\}$ similarly. Then any (surjective) homomorphism $A\to B$ induces a (surjective) homomorphism $f:X\to Y$, which is an isomorphism on all homotopy groups based at the identity. However, $f$ is not an isomorphism on homotopy groups based vertices in $A$ unless the original map $A\to B$ was a weak equivalence.

$\endgroup$
  • $\begingroup$ Thank you for your answer. $\endgroup$ – A.G Feb 8 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.