Let $n\mapsto X^n$ be a cosimplicial simplicial set and $X:= \underset{\longleftarrow}{\rm holim}\ X^n$ the homotopy limit. Is the natural map $$ \pi_0(X) \to \underset{\longleftarrow}{\rm lim}\ \pi_0(X^n).$$ surjective?

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    $\begingroup$ It doesn't have to be surjective. That's exactly what the (some of the) differentials in the Bousfield Kan spectral sequence tell you: whether it's surjective or not. $\endgroup$ May 4, 2015 at 10:34
  • $\begingroup$ can you give me a reference for this please? $\endgroup$
    – Amit H
    May 4, 2015 at 11:41

1 Answer 1


This does not have to be surjective. In all cases, $\varprojlim \pi_0(X^n)$ is the same as the equalizer of the two maps $\pi_0 X^0 \rightrightarrows \pi_0 X^1$ (the rest of the diagram is redundant for calculating inverse limits).

Consider the case where $X^n = (\Delta^n)^{(1)}$, the $1$-skeleton of the standard $n$-simplex. The standard face and degeneracy maps make this into a cosimplicial object (a subobject of the cosimplicial $n$-simplex $\Delta^\bullet$). Then $\varprojlim \pi_0 X^n$ is a single point $*$ because all the spaces $X^n$ are path-connected and so the diagram on $\pi_0$ is constant. However, the homotopy limit in this case is empty: a point in the homotopy limit needs (at least) a point $p$ of $X^0$, a path $\sigma$ between its two images in $X^1$, and a triangle in $X^2$ whose boundary consists of the three images of $\sigma$, but this is always impossible no matter which path $\sigma$ you choose.

(This obstruction to getting any points is an obstruction in $\pi_1 X^2$. Bousfield described this in detail in his paper "Homotopy spectral sequences and obstructions".)


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