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  "body": "\n## Statement\n\nLet $G$ and $H$ be [[groups|G5S1M9]] with identity elements $e_G$ and $e_H$, and let $\\varphi \\colon G \\to H$ be a [[group homomorphism|0JYTG5K9Y4]].\nThen $\\varphi(e_G) = e_H$, and $\\varphi(a^{-1}) = \\varphi(a)^{-1}$ for each $a \\in G$.\n\n## Remarks\n\nThe first equation follows from $\\varphi(e_G) \\varphi(e_G) = \\varphi(e_G) = \\varphi(e_G) e_H$ and [[cancellation|0VSNX4P2Q0]].\nThe second follows from $\\varphi(a) \\varphi(a^{-1}) = \\varphi(e_G) = e_H$ and [[the uniqueness of inverses|0D9NDGE9KP]].\nA consequence is that $\\varphi(a^n) = \\varphi(a)^n$ for each integer $n$.\n\n## Explanations\n\n### L1 human\n\nA map that respects products automatically respects the identity and the inverses.\nYou do not need to ask for these two properties in the definition of a homomorphism.\nFor example, the exponential map sends $0$ to $1$ and $-x$ to $1/e^x$.\n",
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