 Solution

Assume $f$ is injective. If two elements $(e,g),(e^{\prime },g^{\prime })\in E^{\prime }$ have the same image under the first projection $E^{\prime }\to E$, then $e=e^{\prime }$. Applying $\pi$ gives $f(g)=f(g^{\prime })$, so injectivity of $f$ gives $g=g^{\prime }$. Hence the natural map $E^{\prime }\to E$ is injective.