- #1
JonnyG
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I don't know if I'm losing my mind or not, but I am reading Artin's "Algebra", and in section 2.6 - isomorphisms, he writes "Corollary 2.5.9. gives us a way to verify that a homomorphism ##\varphi: G \rightarrow G'## is an isomorphism. To do so, we check that ##\mathrm{ker} \varphi = \{1\}##, which implies that ##\varphi## is injective, and also that ##\mathrm{im} \varphi = G'##, that is, ##\varphi## is surjective"
Corollary 2.5.9: A homomorphism ##\varphi: G \rightarrow G'## is injective if and only if its kernel ##K## is the trivial subgroup ##\{1\}## of ##G##.Obviously Corollary 2.5.9 implies that the homomorphism is an isomorphism onto its image, but I do not see how it implies it's an isomorphism onto the co-domain.
Counter-example: Define ##f: \mathbb{R}^+ \rightarrow \mathbb{R}^{\times}## where ##\mathbb{R}^+## is the additive group of ##\mathbb{R}## and ##\mathbb{R}^\times## is the multiplicative group of ##\mathbb{R}##. Obviously ##f## is an isomorphism onto its image (the multiplicative group of positive real numbers), but it's clearly not a surjection.
Did Artin make a mistake or am I missing something painfully obvious?
Corollary 2.5.9: A homomorphism ##\varphi: G \rightarrow G'## is injective if and only if its kernel ##K## is the trivial subgroup ##\{1\}## of ##G##.Obviously Corollary 2.5.9 implies that the homomorphism is an isomorphism onto its image, but I do not see how it implies it's an isomorphism onto the co-domain.
Counter-example: Define ##f: \mathbb{R}^+ \rightarrow \mathbb{R}^{\times}## where ##\mathbb{R}^+## is the additive group of ##\mathbb{R}## and ##\mathbb{R}^\times## is the multiplicative group of ##\mathbb{R}##. Obviously ##f## is an isomorphism onto its image (the multiplicative group of positive real numbers), but it's clearly not a surjection.
Did Artin make a mistake or am I missing something painfully obvious?