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Is there any homomorphism $h$ from $A \rightarrow B$, such that $A$ has an identity $e$ and $h(e)$ is not an identity in $B$?
($A$ and $B$ are Algebras)

Edit: I gave my teacher the same proof given below in the answers, but she argued that it's possible by giving the following example, Let $A = (N,+)$ and $B = (N, .)$ where N is the set of natural numbers and $.$ is the multiply operator, $h: A \rightarrow B$, be a homomorphism such that, $h(a) = 0$ for all $a$ belonging to A, so as 0 is the identity in $A$ and 1 is identity in $B$, but $h(0) = 0$ according to homomorphism, as $h(a) != 1$ , so this proof fails in this case, can you prove why you solution is correct.

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    $\begingroup$ I think your question is unclear. But the answer should be no, since a homomorphism maps the neutral element $e_A$ in A to the neutral element $e_B$ in $B$. $\endgroup$
    – Cornman
    Nov 10, 2017 at 19:46

1 Answer 1

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Let $h$ be a homomorphism between $A$ and $B$. Then $h(xy) = h(x)h(y)$, so $h(x) = h(ex) = h(e)h(x)$, thus $h(e)$ must be an identity in $B$.

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  • $\begingroup$ So are you saying that there is no such example?? $\endgroup$ Nov 10, 2017 at 19:50
  • $\begingroup$ Since $h$ is arbitrary, then .........? $\endgroup$ Nov 10, 2017 at 19:53
  • $\begingroup$ I gave my teacher the same proof, but she argued that it's possible by giving the following example, Let A = (N,+) and B = (N, .) where N is the set of natural numbers and . is the multiply operator, h:A -> B, be a homomorphism such that, h(a) = 0 for all a belonging to A, so as 0 is the identity in A and 1 is identity in B, but h(0) = 0 according to homomorphism, as h(a) != 1 , so your proof fails in this case, can you prove why your solution is correct. $\endgroup$ Nov 14, 2017 at 15:22
  • $\begingroup$ @AndresMejia In this question it is not necessary for A and B to be groups, they are Algebras. Also Due to existence of identities they should be monoids. $\endgroup$
    – Masquerade
    Nov 15, 2017 at 14:06

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