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Let $f\colon A\to B$ be a map of pointed simplicial sets and let $B$ be Kan-fibrant. Let $0\colon A\to B$ denote the map which factorizes over the basepoint of $B$.

Is it true that $f$ is homotopic to $0$ if and only if the composition $$A\xrightarrow{f}B \xrightarrow{\tiny \mbox{canonical}} \pi_0(B)$$ is the map to the basepoint of the pointed set $\pi_0(B)$ which is pointed by the component of the basepoint? I understand $\pi_0(B)$ as the coequalizer of the two boundary maps from $B_1$ to $B_0$.

howard
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1 Answers1

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No. Suppose $A = B$ is connected and $f = \mathrm{id}$; then $f$ is nullhomotopic if and only if $B$ is contractible. (The condition on $\pi_0$ is trivial in this case.)

On the other hand, if $A$ is weakly contractible, then $f : A \to B$ is nullhomotopic if and only if the composite $A \to B \to \pi_0 B$ is the zero map.

Zhen Lin
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