Do Monoid Homomorphisms preserve the identity?

Question

In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: https://en.wikipedia.org/wiki/Monoid#Monoid_homomorphisms. I haven't found an example of the former, so I thought I'd prove the opposite statement.

I believe that I have proved the opposite assertion, based on the proof that I used to show that Group Homomorphisms preserve the identity. Since I don't use any information stating that elements are invertible, I think my proof is still valid.

Let $M, N$ be monoids, and let $f:M\rightarrow N $ be a homomorphism of monoids. Let $m,e_{M} \in M$ be an arbitrary element and the identity in $M$ respectively .

Then: $$f(m) = f(m\cdot e_{M}) = f(m)\cdot f(e_{M})$$ $$f(m) = f(e_{M} \cdot m) = f(e_{M}) \cdot f(m)$$ Thus: $$f(m)\cdot f(e_{M}) = f(e_{M}) \cdot f(m) = f(m), \forall m \in M $$ This seems to imply my assertion. Is there anything wrong with my proof?

Answer 1

As 57Jimmy points out in their comment, you have not proved that the "identity" you have found is the identity of the whole monoid.

Let make this all formal:

If $f:A\rightarrow B$ is a semigroup homomorphism and $A$ and $B$ are monoids then it is not necessarily true that $f(e_A)=e_B$.

As a counter-example, take your favourite monoid $A$ and then attach an identity to obtain a new monoid, $B$. Then the embedding map $A\hookrightarrow B$ is a semigroup homomorphism, but the image of the identity isn't the identity of $B$. For example, take $A=\{e\}$ such that $e^2=e$ and attach an identity $1$ to obtain a new monoid $B$, so $B=\{e, 1\}$ where $1\cdot e=e=e\cdot 1$ and $1^2=1$. Then clearly the monoid $A$ embeds into $B$, but $e$ is not the identity of $B$ (it is in fact the zero, as $1\cdot e=e$, etc.)

Answer 2

The problem is that you have only proved that $f(e_M)$ is an identity for the elements in the image of $f$, not for all the elements in $N$. This is also specified on Wikipedia. So in general, if you do not require it, it is not true that the identity is preserved. Here is a counterexample:

$$(\mathbb{R},*,1) \to (\mathbb{R},*,1), r \mapsto 0.$$

I personally find it strange to require that there is an identity but not that it is preserved by morphisms...

Answer 3

Given an algebraic structure $(A, f_1, f_2, \ldots)$ in the sense of universal algebra morphism are always required to respect all operations (this also includes the constants, as these are modelled as $0$-ary operations). Similar, substructures are closed under all operations, hence they include the same constants as the original algebraic structure.

A monoid has signature $(M, \cdot, 1)$, and respecting $1$ means that the identity should be preserved, similar a group homomorphism is required to preseve inversion. A submonoid is a subset that contains $1$ and so on.

As pointed out by others, we could not get rid of the requirement that the identity should be preserved, i.e. monoids does not form a full subcategory of semigroups. But as you have shown, the image of the identity forms an identity in the image of the morphism.

But sometimes we have some relations between these concepts, particular in the case of groups. Groups form a full subcategory of semigroups, for a proof see here. Similar, a finite subsets of a group is a subsemigroup iff it is a subgroup, this follows by looking at the powers of elements from the subset. So sometimes authors may define group homomorphisms in this weaker sense, or even subgroups just as subsemigroups if they are just concerned with finite groups. Something similar concerns kernels. In general these are relations defined by looking at that elements that are mapped to the same element under a homomorphims, giving rise to a congruence relation and congruence classes, again in the group case such a kernel is fully specified by just giving a single congruence class which form a normal subgroup, and almost every group theorist uses this more restrictive definition of kernel then. Also in a monoid there could be many subsemigroups that form themselve monoids or even subgroups (again look at the powers of some element), but these are not submonoids if they do not contain the identity of the whole monoid. So, sometimes we do not use all properties of the correct definitions, but in your case of monoids we do need all of them.

Answer 4

Your proof just proves that for every element in $\text{Im}(f)$ the element $f(e)$ acts as a unit but it does not imply that this holds for every element of $N$, unless $f$ is surjective.

Hence it is necessary to require that a monoid homomorphism preserves unit.

Allow me to provide a ciunter-example that explain why we need to require in the definition of monoid-homomorphism the unit preservation properties, i.e. why it cannot be deduced.

Consider the the monoid $\mathbb N$ of the natural numbers with addition and let $B$ be the monoid of truth-values, i.e. $B=\{\bot=false,\top=true\}$ with the monoid structure given by the the logical or $\lor$ and having unit the truth value $\bot$.

The mapping $$\begin{align*}f \colon \mathbb N &\longrightarrow B\\ f(n)&=\top \\ \end{align*}$$ is a semigroup homomorphism, i.e. $$f(n+m)=\top=\top \lor \top =f(n)\lor f(m)$$ but it is not a monoid homomorphism, because it does not preserve the unit, $f(0) \ne \bot$.

Since monoids are semigroups it is interesting studying semigroup-homomorphism between them, nevertheless they should notnregarded as the correct notion of morphism for monoid. A good notion of morphism should oreserve all the structure and as the examle above shows semigroup-homomorphisms may fail in doing so.