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. 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?

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    $\begingroup$ Morphisms should always preserve structures. Having a morphing that does not preserve of part of the structure, the identity is highly dubious. It would make $\mathbf{Mon}$ a very weird category for instance. $\endgroup$
    – Lærne
    Commented Jan 29, 2018 at 15:10
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    $\begingroup$ You've misunderstood Hungerford -- he only defines "homomorphism" for semigroups. The quote you refer to is basically the statement "not every semigroup homomorphism between monoids is actually a monoid homomorphism", except Hungerford never seems to define what a monoid homomorphism is (it must preserve the identity) nor what a group homomorphism is (it must preserve the identity and inverses). $\endgroup$
    – user14972
    Commented Jan 29, 2018 at 15:21
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    $\begingroup$ ... a strange feature of the theory of groups is that every semigroup homomorphism between groups turns out to also be a group homomorphism. Because of this, it is (distressingly, IMO) common for introductory texts to define "group homomorphism" to be "semigroup homomorphism between groups". $\endgroup$
    – user14972
    Commented Jan 29, 2018 at 15:23
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    $\begingroup$ @Hurkyl I think you comment should be posted as answer. Maybe it does not exactly answer as it is posed int the title, but this title is based on a wrong interpretation of the text in Hungerford's Algebra. Comments are volatile objects. We could not count on that they will not be removed. $\endgroup$
    – miracle173
    Commented Jan 29, 2018 at 15:53

4 Answers 4


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.)

  • $\begingroup$ What do you mean by "attach an identity"? $\endgroup$ Commented Jan 29, 2018 at 15:04
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    $\begingroup$ @user I've edited my answer to give an example. This is a pretty standard idea in semigroup theory - just attach an element $1$ such that $1\cdot a=a=a\cdot 1$ for all $a\in A$. (So, for example, every semigroup embeds into a monoid in this way.) $\endgroup$
    – user1729
    Commented Jan 29, 2018 at 15:09
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    $\begingroup$ This example is great. Relatedly, I was wondering what are the necessary and sufficient conditions on monoids for all semigroup homomorphisms to preserve the identity, and the answer is that this is true iff the identity is the only element $y\in B$ that is idempotent ($y^2=y$). See this $\endgroup$
    – postylem
    Commented Jan 25, 2023 at 14:20

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...


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.


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.


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