# Homotopy groups of connected components

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).

• $X=S^1\cup\{0\}$ 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? – freakish Feb 8 at 15:39
• $f$ is a homomorphism of simplicial monoids, not just of simplicial sets. – A.G Feb 8 at 17:29
• 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 $x\neq *$. – freakish Feb 8 at 17:31
• @freakish: That doesn't quite work since it won't induce an isomorphism on $\pi_0$. – Eric Wofsey Feb 8 at 17:32
• Thank you anyway for the first comment. – A.G Feb 8 at 17:37

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.