As you saw by your counter example, a Lie group homomorphism does not induce an isomorphism of fundamental groups. But one way to use homomorphisms to determine fundamental groups is through the fact that if $G$ and $H$ are connected with $G$ simply connected and $G \to H$ is a surjective homomorphism with a discrete kernel $K$ contained in the center of $G$, then this map is a covering and the fundamental group of $H$ is isomorphic to $K$. So if you know that $SU(n)$ is simply connected then you can consider the homomorphism
$$
SU(n) \times \mathbb R \to U(n), ~~ (A, t) \mapsto e^{it} A.
$$
This is surjective with kernel isomorphic to $\mathbb Z$ so that $\pi_1(U(n)) \simeq \mathbb Z$.

Though I guess the easiest way to see that $SU(n)$ is simply connected is to use Neal's suggestion of applying the LES in homotopy associated to the fibration
$$SU(n-1) \to SU(n) \to SU(n)/SU(n-1) \simeq S^{2n-1}.$$

But then you might as well compute $\pi_1 U(n)$ directly from the similar fibration
$$U(n-1) \to U(n) \to U(n)/U(n-1) \simeq S^{2n-1}.$$