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The answer is no, and there are plenty of counterexamples. Note that simplicial sets are not relevent here; one can cook up examples with spaces and take their total singular complexes.

For example, there are high dimensional knots $K: S^n \to S^{n+2}$ (i.e., smooth embeddings with $n > 1$) such that the complement $X = S^{n+2} - K(S^n)$ has $\pi_1(X) = \Bbb Z$. A generator is represented by a map $X \to S^1$ which is a both a $\pi_1$- and a homology isomorphism. This will give examples with the exception of your condition on $\pi_2$.

To get the $\pi_2$ condition on the above consider the subclass of those knots such that $n = 2k+1$ is odd and the universal cover $\tilde X$ is a non-trivial Eilenberg Mac Lane space of dimension \pi_j(X) \cong \pi_j(S^1)$for$(k+1)$. j\le k$ and $\pi_{k+1}(X) \ne 0$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing). In The classification was announced by Kearton in the paper

Classification of simple knot case, the map $X \to S^1$ is an isomorphism on homotopy up through dimension $k$. knots by Blanchfield duality. Bull. Amer. Math. Soc. 79 (1973), 952–955

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The answer is no, and there are plenty of counterexamples. Note that simplicial sets are not relevent here; one can cook up examples with spaces and take their total singular complexes.

For example, there are high dimensional knots $K: S^n \to S^{n+2}$ (i.e., smooth embeddings with $n > 1$) such that the complement $X = S^{n+2} - K(S^n)$ has $\pi_1(X) = \Bbb Z$. A generator is represented by a map $X \to S^1$ which is a both a $\pi_1$- and a homology isomorphism. This will give examples with the exception of your condition on $\pi_2$.

To get the $\pi_2$ condition on the above consider the subclass of those knots such that $n = 2k+1$ is odd and the universal cover $\tilde X$ is a non-trivial Eilenberg Maclane Mac Lane space of dimension $(k+1)$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing). An In the simple knot is such that case, the map $X \to S^1$ is an isomorphism an isomorphism on homotopy up through dimension $k$.

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The answer is no, and there are plenty of counterexamples. Note that simplicial sets are not relevent here; one can cook up examples with spaces and take their total singular complexes.

For example, there are high dimensional knots $K: S^n \to S^{n+2}$ (i.e., smooth embeddings with $n > 1$) such that the complement $X = S^{n+2} - K(S^n)$ has $\pi_1(X) = \Bbb Z$. A generator is represented by a map $X \to S^1$ which is a both a $\pi_1$- and a homology isomorphism. This will give examples with the exception of your condition on $\pi_2$.

To get the $\pi_2$ condition on the above consider the subclass of those knots such that $n = 2k+1$ is odd and the universal cover $\tilde X$ is a non-trivial Eilenberg Maclane space of dimension $(k+1)$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing). An simple knot is such that the map $X \to S^1$ is an isomorphism an isomorphism on homotopy up through dimension $k$.