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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: f(1)=1 dangerous''
by djao on 2003-11-19 22:12:31 |
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| I don't see how your claim holds. If f:R->S and g:S->T both satisfy 1->1 then the composition also satisfies 1->1 (1_R necessarily maps to 1_S, and 1_S necessarily maps to 1_T; therefore 1_R necessarily maps to 1_T). If you think this reasoning is wrong, I would very much appreciate knowing why this reasoning is wrong.
Besides, I don't really have the option of removing the requirement, since the standard definition of ring homomorphism DOES include the requirement.
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