$\pi_2(S^2\vee S^1,S^1)\to\pi_2(S^2\vee S^1/S^1)\cong\pi_2(S^2)$ not injective

Question

I want to show that $\pi_1(S^2\vee S^1,S1)\cong 0,$ and that the map $\pi_2(S^2\vee S^1,S^1)\to\pi_2(S^2\vee S^1/S^1)\cong\pi_2(S^2)$ is surjective but not injective. The first part is clear, because $S^1$ is connected, hence$S^2$ is 1-connected and therefore we get by excision $$\pi_1(S^2\vee S^1,S1)=\pi_1(S^2)=0.$$ Also surjectivity is clear because we have that f.a. $[f]\in\pi_2(S^2)$ the image under $\pi_2(S^2)\to\pi_2(S^2\vee S^1,S^1)$ is mapped onto $[f].$
Unfotunately I don't get why the map is not injective, do you have an proof for this?