do morphisms of binary operations have to be morphism of units?

Question

? For monoids $(M,b,u)$ and $(M',b',u')$, show by an example that a morphism $f\colon (M,b)\to (M',b')$ of binary operations need not be a morphism of units.

This is a homework problem I got. I think it doesn't make sense, because if $u$ is the unit of the binary operation $b$ on $M$, then by definition of morphism, for all $x$ in $M$, $f(x\,b\,u)=f(x)$ and $f(x\,b\, u)=f(x) \,b'\, f(u)$, so $f(x)=f(x)\, b'\, f(u)$ for all $x$, then by definition of unit, $f(u)$ is the unit of the binary operation $b'$ on $M'$.

So this means that all morphisms of binary operations have to be morphisms of units at the same time. I don't understand why this problem wants me to think of an example that it's not so.

Have I misread this problem, or is it just wrong?

Thanks

By the way, is it OK if I just find an $f(x)$ that's not a surjection, so that $f(x) \,b'\, f(u)=f(x)$ does not work for all the elements in $M'$?

Consider for instance the map from $(\mathbb{N},\cdot)$ -- i.e., the set of non-negative integers under multiplication -- to itself which sends every $n$ to $0$. — Pete L. Clark, Sep 26 '11 at 05:51
but does that mean my proof is wrong? can you explain how it is wrong? — Scharfschütze, Sep 26 '11 at 05:59
wait....if 0 is not in N, how is f a function from (N, ) to (N, ) then? — Scharfschütze, Sep 26 '11 at 06:00
$@$Scharfschütze: I said that $\mathbb{N}$ denotes the set of non-negative integers...so it includes zero. (Zero is not negative!) And yes, the function I gave you is a (homo)morphism of monoids. — Pete L. Clark, Sep 26 '11 at 06:09
@Pete L. Clark: Oh yeah cos I thought you meant the natural numbers. Thanks — Scharfschütze, Sep 26 '11 at 06:27
@Pete L. Clark: but I still don't see how my proof is wrong...can you help me with that? — Scharfschütze, Sep 26 '11 at 06:40
Your proof only shows, that $f(u)$ is the unit on the image of $f$ and as $f$ is not necessarily surjective this in general isn't the unit of the whole monoid as the example shows. On the other hand, if $f$ is surjective, then $f$ necessarily maps unit to unit. — Aleš Bizjak, Sep 26 '11 at 06:46

score 3 · Accepted Answer · answered Sep 26 '11 at 13:31

As has been pointed out, your argument shows that $f(u)$ is a unit of $f(M)$, not that $f(u)$ is a unit of $M'$. The argument suffices for surjective maps, but not for non-surjective maps.

An easy example is to take $f\colon(\mathbb{N},\times,1)\to(\mathbb{N},\times,1)$, with $f(n)=0$ for all $n$ (my naturals have $0$ in them; if yours don't, then just take he nonnegative integers instead.

Another example is to take any monoid with a zero element, say $(\mathbb{Z},\times, 0)$, and consider $M=(\mathbb{Z},\times,1)$ and $M'=(\mathbb{Z}\times\mathbb{Z},\times,(1,1))$, and let $f$ be the inclusion into the first coordinate, $f(n) = (n,0)$.

You can probably now cook up lots of examples along similar lines. All you need is an idempotent (an element $e$ such that $e^2=e$) which is not the unit of $M'$.

So, two downvotes, within a minute of one another, with no explanation on either... — Arturo Magidin, Sep 26 '11 at 15:54

do morphisms of binary operations have to be morphism of units?

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