Let $X=|\Delta|$ be the geometric realization of an abstract simplicial complex $\Delta$. Let $k$ be a field. Assume that $X$ is path connected .
Consider the following two conditions:
(1) $\pi_i (X)=0, \forall 1\le i\le n$
(2) $\tilde H_i(X,k)=0, \forall 0\le i\le n$
I know that (1) implies (2) by Hurewicz theorem and Universal coefficient theorem. My question is : Does (2) imply (1) ? If not, then would some other vanishing result on (reduced) Homology imply (1) ?
The Poincare Homology sphere has $\tilde H_0$, $H_1$ and $H_2$ trivial, yet has non-trivial $\pi_1$. But the Abelianisation of $\pi_1$ is trivial. I suspect that "(2) implies (1)" is far too optimistic, even if you replace $\pi_1(X)$ by its Abelianisation in (1).
If we use $\mathbb Z$-coefficients, then the Hurewicz theorem tells us that (2) implies (1) provided $X$ is simply connected. If it is not, then there are counterexamples (see Lord Shark the Unknown's answer).
If we use coefficients in a field $k$, then it is possible that (2) is satisfied, but $\tilde{H}_i(X) \ne 0$ for some $i \le n$. As an example take the space $X$ obtained by attaching to $S^2$ a cell $D^3$ via map $f : S^2 \to S^2$ of degree $2$. This space is simply connected and has $\pi_2(X) = \mathbb Z_2$. By the Hurewicz theorem $H_2(X) = \mathbb Z_2$ and by the universal coefficient theorem $H_2(X,\mathbb Q) = 0$. See An abelian group A is torsion iff A ⊗ Q = 0.
This example shows that (for $k = \mathbb Q$) even for simply connected $X$ (2) does not imply (1).
However, there exists a "rational Hurewicz theorem" which allows to get information about $\pi_i(X) \otimes \mathbb Q$. See https://en.wikipedia.org/wiki/Hurewicz_theorem.