Let $\phi$ be a bijective function. Can we conclude that two monoids $(M_1,.)$ and $(M_2,∗)$ are isomorphic?

Condiser two monoids $$(M_1,.)$$ and $$(M_2,*)$$ with identity elements $$e_1$$ and $$e_2$$ and a bijective function $$\phi$$ which has the property $$\phi(a.b)=\phi(a) * \phi(b)$$ for all $$a,b \in M_1$$

can we conclude that $$(M_1,.)$$ and $$(M_2,*)$$ are isomorphic?

My thoughts:

from this A homomorphism between two monoids $$(M_1,.)$$ and $$(M_2,*)$$ is a function f : $$M_1$$$$M_2$$ such that

1. $$f(a.b)=f(a) * f(b)$$ for all $$a,b \in M_1$$
2. $$f(e_1)=e_2$$

and then a bijective monoid homomorphism is called a monoid isomorphism.

I'm saying if $$f$$ is bijective we need to show for all $$c \in M_2$$ : $$c*f(e_1)=f(e_1)*c=c$$ and then we can conclude $$f(e_1)=e_2$$ and the second condition is not nesseccery for being isomorphic.

Consider $$c \in M_2$$. beacuse $$f$$ is bijective so there is $$x \in M_1$$ which $$f(x)=c$$ so $$c*f(e_1)=f(x)*f(e_1)=f(x.e_1)=f(x)=c$$ and by doing the same for $$f(e_1)*c$$ we get $$c*f(e_1)=f(e_1)*c=c$$ and as I said $$f(e_1)=e_2$$. so the second condition is not nesseccery for being isomorphic. and only $$f$$ being bijective is enough.

Is the conclusion I made correct?

• @Shaun Thank you! Oct 18, 2020 at 8:21
• @Joshua Cole There is an important conceptual point to be made: what we call (definitionally) a monoid isomorphism is not a bijective monoid morphism, but a monoid morphism which happens to admit a(nother) monoid morphism as an inverse. It is a matter not of definition but of subsequent characterisation that a monoid morphism is an isomorphism if and only if it is bijective, and this phenomenon only occurs for algebraic structures (universal algebras) and not in other categories such as topological spaces or relational systems.
– ΑΘΩ
Oct 18, 2020 at 8:25
• @Joshua Cole But of course, I apologise for the terse style. What I meant is that the phenomenon of "isomorphisms coinciding with bijective isomorphisms" is valid for algebraic structures (groups, rings, modules etc) but not for the category of say ordered sets. There exist quite elementary examples of bijective morphisms of ordered sets (the morphisms in this category are the isotonic or increasing maps) which are however not isomorphisms. The same goes for topological spaces: there exist continuous bijections which however fail to be homeomorphisms (isomorphisms).
– ΑΘΩ
Oct 18, 2020 at 9:23
• @Joshua Cole In order to assimilate this difference in how morphisms behave in various categories, a simple introduction to category theory -- combined with some exercises that illustrate what the general abstract notions become in the concrete instances of sets, groups, monoids, topological spaces etc -- would be more than sufficient. You could consider "Categories for the working mathematician" of Saunders MacLane as a first introduction.
– ΑΘΩ
Oct 18, 2020 at 10:27
• @Joshua Cole Concrete experience with classical categories such as that of sets, semigroups, rings, modules, topological spaces, ordered sets etc would certainly help but it is not an absolute prerequisite. To help you form some measure of such experience, may I invite you to try and produce examples of bijective morphisms which however are not isomorphisms in whichever of the following categories you might be more familiar with: ordered sets, topological spaces or simple graphs. If you would like to ask more questions, feel free to do so (we could take this discussion over to the chat room).
– ΑΘΩ
Oct 18, 2020 at 13:52

Yes, you have the right idea. To clarify the language slightly, you are asking if an isomorphism between $$M_1$$ and $$M_2$$ as semigroups is automatically an isomorphism as monoids. It is easy to check that a bijective semigroup (resp. monoid) homomorphism is automatically an iso, so your question is the same as this one.

I will say up-front that your computation is correct, and so is your conclusion. I will also say that there is some extremely interesting mathematics lying just under the surface. If you'll indulge me, I would love to share it with you ^_^

First let's note that some bonus condition is crucial. Consider $$M_1 = M_2 = (\mathbb{Z}^2, \times)$$, with componentwise multiplication. Then the map $$f(a,b) = (a,0)$$ is easiliy seen to be a semigroup homomorphism, but $$f(1,1) = (1,0) \neq (1,1)$$ so the identity is not preserved.

Notice, however, that this is really a counterexample by a technicality. $$f(1,1) = (1,0)$$ is an identity for the image $$f[\mathbb{Z}^2]$$. After all $$(a,0)(1,0) = (a,0) = (1,0)(a,0)$$. It is only for elements not in the image, like $$(0,b)$$ that we notice $$(1,0)$$ fails to be an identity. At the risk of outing myself as a logician, I would love to talk about the model-theoretic content in this observation:

The property of "being an identity" is expressible in the language of semigroups. Let $$\varphi(x,y)$$ be the formula $$yx = x \land xy = x$$. Then $$e$$ is an identity if and only if $$\forall x . \varphi(x,e)$$ is true. Now since $$\varphi$$ is "positive" (in the sense that it doesn't have any negations), it is preserved by arbitrary homomorphisms. So if $$\varphi(x,y)$$ is true in $$M_1$$, then $$\varphi(f(x),f(y))$$ will be true in $$M_2$$.

Notice this almost the same as saying that if $$\forall x . \varphi(x,e)$$ is true in $$M_1$$, then $$\forall x' . \varphi(x',f(e))$$ is true in $$M_2$$. The problem is the range of the quantifiers. The first quantifier ranges over the elements of $$M_1$$, whereas the second ranges over all the elements of $$M_2$$. Of course, we (in general) have no control over the parts of $$M_2$$ outside the image of $$M_1$$, so it makes sense that this "strong" universal quantifier might fail to be true. But we are guaranteed that $$\varphi(x',f(e))$$ is true when we promise to look only at $$x'$$ in the image of $$M_1$$.

So, in the special case that $$f$$ is surjective, we can see how to proceed. If $$f$$ is surjective, then every element of $$M_2$$ is in the image of $$M_1$$. So we really can put the universal quantifier in front, and the property of "being an identity" is preserved.

This gives us a slightly stronger claim than what you wanted: It suffices that the semigroup homomorphism be surjective. And if you look at the computational proof that you gave, you only used surjectivity when you concluded that the identity was preserved!

The reason to go on this long diversion is to give you a tool to see not only that this is true, but to see how it might be obviously true. The property of "being an identity" is definable in the language of semigroups, and isomorphisms preserve all first order formulas. So, in particular, the identity gets mapped to an identity under a semigroup isomorhpism. This kind of argument is extremely flexible, and I hope it serves you well going forwards!

I hope this helps ^_^

• nice answer!  Oct 18, 2020 at 8:29
• @HallaSurvivor On an even more general level, given a unitary magma $(A, \cdot)$, if there exists a surjective magma morphism $f \colon A \to B$, it follows that the target magma $(B, \cdot)$ is also unitary and the relation $f(1_A)=1_B$ necessarily holds (in other words, $f$ is automatically a morphism in the appropriate category of unitary magmas).
– ΑΘΩ
Oct 18, 2020 at 8:31
• @ΑΘΩ - That actually follows from this same line of reasoning. We never use associativity in $\varphi$, so this exact same proof shows that any identity in a magma is sent to an identity by a surjective magma homomorphism. It's nice to point this out explicitly, though ^_^ Oct 18, 2020 at 8:35
• @HallaSurvivor Thank you very much! I'm a first-year so there were some points that I couldn't follow but even in a situation like that I enjoyed your answer and it was VERY nice, again, thank you. Oct 18, 2020 at 8:42
• @HallaSurvivor Indeed it does, as you very well put it. Permit me to make just one more remark, namely that any two identities of a magma must necessarily coincide, so if a magma is unitary it then has a unique identity, which justifies the use of the definite article in the syntagm "the identity of a (unitary) magma". A pleasure chatting, sir.
– ΑΘΩ
Oct 18, 2020 at 9:28