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I'm familiar with the definition (inverses and bijections, preserving operations) in the context of groups and vector spaces and do know the formal definitions of morphisms as a whole.

However, what I'm looking for is a mathematical justification and a sharp intuition for the purpose of defining homomorphisms and isomorphisms. I find it quite challenging to picture them and find myself unable to convince myself when I'm trying to explain those concepts.

Right now, the only analogy I can make is one for the isomorphisms in the context of groups, namely I like to think about the isomorphism between $\mathbb{Z}_2$ with the addition and the set {$-1,1 $ } with multiplication where we map $0$ to $1$ and $1$ to $-1$, this way adding two even numbers or two odd numbers is analogous to multiplying elements of {$-1,1$ } , although when I try to abstract this example further I hit some brick walls :

  • Making Sense of $f(a) f(b)=f(a \star b)$ where juxtaposition is the operation of a group $G$ and the star is the operation of a group $H$
  • I cannot see how the structure of the group $G$ is being preserved in $H$.

I would like an answer giving intuition about those concepts rather than their formal definitions. Note that I already consulted other threads concerning this question, so this post aims to look at other perspectives in hope to understand better.

As i have already poined the out , i've already seen and read the linked thread multiple times but did not personally get a satisfactory answer , what i'm asking for is to shed some light upon how the preserving of group structure is being translated through $f(a) f(b)=f(a \star b)$ .

Adam Boussif
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    For finite groups, an isomorphism tells you how to relabel the Cayley table for $G$ into the table for $H$, so they have the "same" structure. – Randall Sep 04 '23 at 11:26
  • Namely, let $f\colon G\to G'$ a bijection. With reference to the following diagram: $$\begin{array}{c} G & \stackrel{f}{\longrightarrow} & G' \ \downarrow{\mathfrak c} & & \downarrow{\mathfrak{c'}} \ \operatorname{Sym}(G) & \stackrel{\psi_f}{\longleftarrow} & \operatorname{Sym}(G')
    \end{array} $$ where $\mathfrak c$ and $\mathfrak c'$ are Cayley's embeddings and $\psi_f\colon\operatorname{Sym}(G')\to\operatorname{Sym}(G)$ is defined by $\sigma\mapsto f^{-1}\sigma f$, the following result holds true: $$\space\forall g,h\in G: f(gh) = f(g)f(h) \iff \mathfrak c=\psi_f\mathfrak c'f$$     – citadel Sep 04 '23 at 11:30
  • In words, you seem not trusting that the operation-preserving property of a bijection between two groups rightly conveys the notion of "having the same structure". And I can share that unconfortability with you. So, if you get aware that a group's structure live in the group 's symmetric group (Cayley's theorem), then perhaps the commutativity of the above diagram gives you a more accepable notion (definition, actually) of "isomorphicity". But then, as a lemma, this is equivalent to the usual operation-preserving property. – citadel Sep 04 '23 at 12:00
  • @citadel: The Question here is broad, and perhaps the OP is looking for an "intuition" that applies beyond group theory. But it may be that your Comments meet the OP where they are and give a bridge to a broader perspective. – hardmath Sep 04 '23 at 12:15
  • @citadel I 'll make more research on the subject by taking your advice and looking at cayley's theorem . However as -hardmath said i am indeed looking for a general "intuition" for homorphisms and isomorphisms that could be generlized to all algebraic structures – Adam Boussif Sep 04 '23 at 12:44
  • Fair enough, though reading and reading again the post, its tags, and the answer, I had the impression that you were group-focused. – citadel Sep 04 '23 at 14:03
  • Regarding the $f(a)f(b)=f(a\star b)$ thing, it may be easier to understand in this form: $f(a) f(b) = f(c)$ if and only if $a\star b=c$. The two formulations are equivalent, as you should show. – MJD Sep 05 '23 at 21:41

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The basic intuition behind homomorphisms is the following: if $G$ and $H$ are two algebraic structures of the same type (e.g. groups, vector spaces, etc.), and $\varphi:G\to H$ is a homomorphism, then a homomorphism is a "structure-preserving" map. Now, what does "structure-preserving" mean? This term is a little loose, and it easiest to understand it with examples:

  • Every group has a binary operation, an identity element, and inverses for every element. These three "operations" comprise the structure of the group. If $\varphi:G\to H$ is a group homomorphism, then $\varphi$ "preserves" the structure in the sense that $\varphi(a)\varphi(b)=\varphi(ab)$, $\varphi(e_G)=e_H$, and $\varphi(a^{-1})=\varphi(a)^{-1}$. It turns out that simply requiring the first condition ensures that the other two hold, hence the definition given in textbooks. These three equations have a common thread to them which I will try to distill here. For each $a\in G$, it helps to think of $\varphi(a)$ as the "corresponding element" to $a$: the way in which $a$ interacts with other elements of $G$ is similar to the way $\varphi(a)$ interacts with other elements of $H$. For instance, if an equation $ab=c$ holds in $G$, then replacing $a$, $b$, and $c$ with each of their corresponding elements gives us the equation $\varphi(a)\varphi(b)=\varphi(c)$, which holds in $H$. Similarly, if $a^{-1}=b$, then $\varphi(a)^{-1}=\varphi(b)$. Finally, a group homomorphism sends the identity of $G$ to the identity of $H$.
  • The situation with vector spaces is similar, except that there is more structure to be preserved than with groups. A vector space homomorphism, which for historical reasons tends to be called a "linear map", preserves all the operations that a group homomorphism does, but it also preserves scalar multiplication. It turns out that simply requiring that $\varphi(u+v)=\varphi(u)+\varphi(v)$ and $\varphi(\lambda v)=\lambda\varphi(v)$ suffices. (Warning: in a vector space, the group operation is written using additive notation rather than multiplicative notation, and scalar multiplication is written using multiplicative notation; don't let this confuse you.)

An isomorphism between $G$ and $H$ is an invertible homomorphism: that is, it is a homomorphism from $G$ to $H$ which has an inverse that it also a homomorphism. In the case of algebraic structures such as groups and vector spaces, every bijective homomorphism is an isomorphism. (In more exotic categories, such as the category of topological spaces, this is not so: a bijective continuous map does not necessarily have a continuous inverse.)

If there exists an isomorphism between $G$ and $H$, then they are said to be isomorphic. Isomorphic structures are often regarded as "the same", but this is a more subtle concept than you might think. In Michael Artin's Algebra, he gives the following analogy:

When $\varphi:G\to G'$ is an isomorphism, we can make a computation in either group. then use $\varphi$ or $\varphi^{-1}$ to carry it over to the other. So, for computation with the group law, the two groups have identical properties. To picture this conclusion intuitively, suppose that the elements of one of the groups are put into unlabeled boxes, and that we have an oracle that tells us, when presented with two boxes, which box contains their product. We will have no way to decide whether the elements in the boxes are from $G$ or from $G'$.

It is instructive to think hard about why the formal definition really does align with the "box analogy". Let me know if you have any questions.

Joe
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  • Thanks for this great answer ! I am still struggling a bit with the "box analogy" although i'm having a clearer picture now whcih is already a good start . I'll try to explain the concept as an auto-Test , please do let me know if i fail at doing so – Adam Boussif Sep 04 '23 at 13:29
  • Basically the structure is preserved in the sense that elements of H act in the same way as elements of G and under the corresponding operation of G ! , for instance taking an element "a" of G and it's " corresponding element" $f(a)$ we should expect that $f(a)$ acts similar to a with elements of H , so if a.b=c in G then written litterally the corresponding element of a and the correspondant element of b under the corresponding operation of G spits out the correspondant element of c which translates to $f(a)f(b)=f(a \star b )$. – Adam Boussif Sep 04 '23 at 13:32
  • Similarly the corresponding indentity of G should be the identity of H or $f(e)=e'$ , and the corresponding inverse of an element a in G should be the inverse of the corresponding element a in H which means $f(a^{-1})=f(a)^{-1}$ and it is a group isomorphism if we can not distinguish between the interactions in neither groups so the product in H - $f(a)f(b)$ is indistinguishable from the product in G - $a\star b $ . How did i do ? – Adam Boussif Sep 04 '23 at 13:37
  • @AdamBoussif: That is pretty much correct. If $\varphi:G\to H$ is a group homomorphism, then for each equation $G$ that holds, there is a corresponding equation which holds in $H$. However, if $\varphi$ is not injective, then there could be multiple elements of $G$ which have the same the same corresponding element. – Joe Sep 04 '23 at 14:03
  • so we would require to have an element in H that uniquely corresponds to a , that way if $f(a)=f(b)$ it implies $a=b$ along with surjectivity the two groups have the same order and correspondant elements and operations meaning they are the same from a group theoretical perspective ? – Adam Boussif Sep 04 '23 at 14:07
  • @AdamBoussif: If $\varphi$ is an isomorphism, then there is a perfect correspondence between the equations that holds in $G$ and the equations that hold in $H$. – Joe Sep 04 '23 at 14:07
  • In the box analogy, we cannot work out whether the group is $G$ or $G'$ because if it is possible that the boxes contain $a,b\in G$, and their product is $c$, it is equally possible that the boxes contain $\varphi(a),\varphi(b)\in H$, and their product is $\varphi(c)$; and if the boxes contain $x,y\in H$, and their product is $z$, then it is equally possible that they contain $\varphi^{-1}(x),\varphi^{-1}(y)$, and their product is $\varphi^{-1}(z)$. – Joe Sep 04 '23 at 14:07
  • @AdamBoussif: To answer your final question, yes, we would say that they are the "same" from a group-theoretical perspective. The only way to distinguish isomorphic groups is to look at what the boxes actually contain, and that is irrelevant to group theory: what matters is how the elements interact with each other, rather than what they "are". – Joe Sep 04 '23 at 14:08
  • Thanks for your awesome answers , i believe now i have a better understanding of homomorphisms and isomorphisms as well as their purpose as far as i'm concerned , it was how we think of $f(a)$ as corresponding element of a that made it click , such a simple and natural interpretation yet i couldn't think of it . Thanks again ! – Adam Boussif Sep 04 '23 at 14:11
  • @AdamBoussif: No problem! Did the box analogy end up making sense to you? – Joe Sep 04 '23 at 14:22
  • Yes , it does ! – Adam Boussif Sep 04 '23 at 14:26
  • @AdamBoussif: It seems likely that this question will eventually be closed and/or deleted. This is not entirely unreasonable, since I do think that questions similar to yours have been asked many times on this site. I would suggest that you take a screenshot of my answer on your computer if you want to refer to it in the future. – Joe Sep 05 '23 at 22:36