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I am trying to find a reference or an explanation for the result that if $X_\bullet$ is a simplicial space with $X_0$ a point then $\pi_1(|X_\bullet|)$ is the free group on $\pi_0(X_1)$ $/ \sim$, where $\sim$ is the relation that $\partial_1(x) = \partial_2(x) \partial_0(x), \forall x\in \pi_0(X_2) $. I think alternatively if I could show that $\pi_1(\sum X_1)$ is the free group on $\pi_0(X_1)$, then this would be sufficient but this doesn't seem to be true as this would imply that the fundamental group of the reduced suspension is always a free group which might not be true (at least to my knowledge). Any suggestions or references would be great.

treutm14
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I don't know about simplicial spaces. For simplicial sets this is first due to Kan, A combinatorial definition of homotopy groups. See section 19. (Wikipedia discusses the case of simplicial complexes, and the ideas should be similar.)

John Palmieri
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  • Thanks for the response, I agree that the answer is probably very closely related to this and probably looks close to identical in the case of a simplicial set with $X_0$ a point, I imagine (if $X_1$ was a set not a space) you would probably just replace $\pi_0(X_1)$ with the free group on $X_1$ and the relations being held for all elements of $X_2 $(I think). Would you know why we would replace $X_1$ with $\pi_0(X_1)$ in the case where X is not a simplicial set but a simplicial space? – treutm14 Nov 30 '23 at 08:03
  • I'm sorry, I don't know. – John Palmieri Nov 30 '23 at 21:02