I just need a check on this. I feel that the proof is weak, but I want to make sure.
Let $\phi: X \rightarrow Y$ be a continuous map of topological spaces. Define $\phi_\ast: \pi_1 (X, x) \rightarrow \pi_1 (Y, \phi(x))$ such that $\phi_\ast ([f]) = [\phi \circ f]$
Consider $\phi_\ast([f]\cdot[g]) = [\phi \circ ([f]\cdot [g])]$. Since $\phi$, $f$, and $g$ are continuous maps, we know that $[\phi \circ ([f]\cdot[g])] = [\phi\circ [f]] \cdot [\phi \circ [g].$
Since $(\phi\circ [f]) \cdot (\phi \circ [g]) = \phi_\ast([f]) \cdot \phi_\ast([g])$, we have shown that $\phi_\ast$ is a group homomorphism.