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The definition of monoid homomorphisms requires that $\eta(1)=1'$. However, I doubt whether this condition is superfluous, for

$\forall x \in M, \eta(x) = \eta(x*1) = \eta(x)\eta(1)$

and

$\forall x \in M, \eta(x) = \eta(1*x) = \eta(1)\eta(x)$

so we may infer that $\eta(1)$ acts as $1'$ in $\eta(M)$, utilizing the condition $\forall a, b \in M, \eta(ab)=\eta(a)\eta(b)$ only.

So is it really necessary to emphasize this condition?

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2 Answers

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No, preservation of multiplicative identity cannot be deduced from the other axioms.

Even for monoids as rigid as the multiplicative monoids of rings, it is not true.

For example, if $e$ is a central idempotent (that is, $e^2=e$) which is not $0$ or $1$ in a ring $R$, then $eRe$ is a ring with identity $e\neq 1$, but it is also a subring of $R$. Then the inclusion map $eRe\rightarrow R$ is a monomorphism of rings, but the identity is not preserved.

Lord_Farin made a very good point though in his solution that if the homorphism is onto, then it follows that it preserves identity, and in general the image of the identity is the identity of the image (but possibly not the identity of the entire codomain).

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True it is that the image of the identity is the identity of the image. Maybe my question should be, if we drop this axiom, what will happen? –  Ezra Oct 19 '12 at 15:08
    
Dropping the axiom redefines the morphisms of the category you're discussing, and that can sometimes impact what you can say about the category. Without the axiom, you will have more morphisms than you had with the axiom in place. –  rschwieb Oct 19 '12 at 16:30
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Generally, the condition can be left out when $\eta$ is surjective. However, otherwise, for $\eta: X \to Y$, there may be $y \in Y$ such that $\eta(1_X)y \ne y$.

For example, consider the inclusion $n \to (n,0)$ of $(\Bbb Z,\cdot)$ into $(\Bbb Z^2, \cdot)$ (coordinatewise $\cdot$, that is).

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