Product with Inverse on Homomorphic Image is Group Homomorphism/Mistake

Source Work
{{{{BookReference|A Course in Group Theory|1996|John F. Humphreys}}: Chapter $8$: The Homomorphism Theorem:
 * Exercise $3$

Mistake

 * Suppose that $H$ is an abelian group and let $\vartheta: G \to H$ be a homomorphism. Define a map $\phi: G \times G \to H$ by
 * $\map \phi {g_1, g_2} = \map \phi {g_1} \map \phi {g_2}^{-1}$.''
 * Prove that $\phi$ is a homomorphism.

Correction
The condition is wrong.

It should read:


 * $\map \phi {g_1, g_2} = \map \vartheta {g_1} \map \vartheta {g_2}^{-1}$