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 $(k+1)$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing). In the simple knot case, the map $X \to S^1$ is an isomorphism on homotopy up through dimension $k$.

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 $\pi_j(X) \cong \pi_j(S^1)$ for $j\le k$ 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). The classification was announced by Kearton in the paper

Classification of simple knots by Blanchfield duality. Bull. Amer. Math. Soc. 79 (1973), 952–955

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 $k$.

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 $(k+1)$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing). In the simple knot case, the map $X \to S^1$ is an isomorphism on homotopy up through dimension $k$.