This might be an incredible easy question -- since any reference I've found state it as obvious -- but anyway:
Given a group $G$, I can construct the full group-$C^*$-algebra $C^*(G)$ be completing $L^1(G)$ in the norm
$$||f||=\sup\{||\pi(f)|| | \pi \text{ is a $*$-representation of } L^1(G)\}$$
as well as the reduced group $C^*$-algebra $C_r^*(G)$ by completing $L^1(G)$ in the norm
$$||f||_r = ||\lambda(f)||_{B(L^2(G))}$$
with $\lambda$ being the left regular representation. Obviously, $||f||_r\leq ||f||$ for any $f\in L^1(G)$, so the identity map on $L^1(G)$ extends to a continuous map $C^*(G)\to C_r^*(G)$. Why is this map -- commonly referred to as the 'canonical surjection' -- surjective?
First note that $L^1(G)$ is dense in both the full and reduced group $C^*$-algebras (by definition). Thus the image of the map from $C^*(G)$ to $C^*_r(G)$ is dense in $C^*_r(G)$. However the image of a $C^*$-algebra under a homomorphism is closed and thus must be all of $C^*_r(G)$. So the map is indeed a surjection.
To see that the homomorphic image of a $C^*$-algebra is indeed closed, I direct you to this answer: https://math.stackexchange.com/a/435105/8580, or you can see Theorem 4.1.9 in Volume I of Kadison-Ringrose.