Isomorphic and Homomorphic

Question

I was wondering if someone could explain the ideas behind isomorphisms and homomorphisms and the difference between them. I understand that a linear map is an isomorphism if it is bijective, and that an isomorphism is a bijective homomorphism but I don't fully understand what a homomorphism actually is.

For example, what does it mean to say that $Hom_{\mathbb{C}}\left(\mathbb{C}^2,\mathbb{C}^3\right)\simeq MAT_{\mathbb{C} }\left(3,2 \right)$ where $MAT_{\mathbb{C}}$ is the $3\times2$ matrix with complex entries.

After some research, I think that this example is saying that $Hom_{\mathbb{C}}\left(\mathbb{C}^2,\mathbb{C}^3\right)$ is essentially a perfect approximation of $MAT_{\mathbb{C} }\left(3,2 \right)$ however I am struggling to understand what the notation $Hom_{\mathbb{C}}\left(\mathbb{C}^2,\mathbb{C}^3\right)$ actually means.

Note: I haven't studied a course in group theory and therefore have little knowledge on it, this is for a Linear Algebra Course

Answer 1

That statement is something of a mathematical quagmire because there are lots of algebraic structures floating around, and the maps between them aren't explicitly "typed" (i.e., it's not exactly explicit what kinds of structures the homomorphisms are between)!

The notation $\operatorname{Hom}_{\Bbb C}(\Bbb C^2, \Bbb C^3)$ is just the set of all $\Bbb C$-linear maps from $\Bbb C^2$ to $\Bbb C^3$ (the subscript $\Bbb C$ is what tells us the maps are $\Bbb C$-linear). Recall that a $\Bbb C$-linear map $T: \Bbb C^2 \to \Bbb C^3$ is just a map that

• "respects addition," so that $T(x + y) = T(x) + T(y)$ for vectors $x, y \in \Bbb C^2$, and

• "respects (complex) scalar multiplication" so that $T(cx) = cT(x)$ for scalars $c \in \Bbb C$ and vectors $x \in \Bbb C^2$.

Another term for a linear map is a vector space homomorphism, hence the notation $\operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3)$. Essentially, as other answers have pointed out, homomorphisms are just maps between algebraic structures (in this case, vector spaces) that "respect" the relevant operations (in this case, vector spaces are spaces in which we can add vectors together and multiply them by scalars, and each operation needs to be "respected").

Now, the sets $\operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3)$ and $\operatorname{Mat}_\Bbb C(3, 2)$ can themselves be thought of algebraic structures (they are groups under composition/multiplication) but I would not worry about that! At least for now.

I would simply view the '$\simeq$' in $\operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3) \simeq \operatorname{Mat}_\Bbb C(3, 2)$ as the claim that these two sets are in bijection: If you have a linear map $T\colon \Bbb C^2 \to \Bbb C^2$, there's an easy way to write down a $3 \times 2$ matrix $M_T \in \operatorname{Mat}_\Bbb C(3, 2)$: Just make the $i$th column of $M_T$ the image $T(e_i)$ of the standard basis vector $e_i \in \Bbb C^2$.

Now, if we want to consider the '$\simeq$' in $\operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3) \simeq \operatorname{Mat}_\Bbb C(3, 2)$ as some kind of isomorphism of algebraic structures (not just between sets, i.e., a bijection), we would then need to worry about whether the relevant operations are preserved. If we're composing maps $S, T \in \operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3)$, we'd need to verify that the corresponding matrices $M_S, M_T \in \operatorname{Mat}_\Bbb C(3, 2)$ multiply the way we'd expect. That is,

$$M_T M_S = M_{T \circ S}.$$

If you show this, then you'll have shown that the groups $\operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3)$ and $\operatorname{Mat}_\Bbb C(3, 2)$ are isomorphic (provided you've already shown they're in bijection). The homomorphism could be made concrete as a map $M \colon \operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3) \to \operatorname{Mat}_\Bbb C(3, 2)$ that sends a $\Bbb C$-linear map $T \in \operatorname{Hom}_\Bbb C (\Bbb C^2, \Bbb C^3)$ to the matrix $M_T \in \operatorname{Mat}_\Bbb C(3, 2)$ described above.

But unless your linear algebra class is way fancier than mine (and it very well may be), I would assume, not having done much group theory, that the statement is just pointing out that the two sets are in bijection, possibly "to be expanded upon later" or "plus other good stuff."

Answer 2

I'm going to phrase this in terms of group theory, but keep in mind that these concepts generalize (with the appropriate hypotheses) to rings, fields, modules, vector spaces, etc.

One of the deep insights of modern math is that if we want to learn about some class of things (e.g. groups), we should also look at functions which map groups to groups. This is a categorical point of view, and motivates the study of abstract categories.

A homomorphism is a map between groups which preserves the group structure, and an isomorphism is a bijective homorphism. Isomorphism means two groups are basically the same. For example, I assume you have been introduced to both dihedral and symmetric groups. I encourage you to convince yourself that the dihedral group of the triangle is isomorphic to the symmetric group on three letters. When we say two groups are isomorphic, it essentially means they are the same object with different names. Therefore, if we want to classify groups, we need only do so up to isomorphism.

Homomorphisms are more subtle, since in general they map between non-isomorphic groups. However, we can learn quite a bit about a group by studying a homomorphism of it into some other group.

For example, the kernel of a homomorphism is a the elements in the grpup which are sent to the identity. I encourage you to prove this as an exercise for your own understanding, but it turns out that the kernel of a homomorphism is a subgroup. In fact, it is a normal subgroup (invariant under conjugation).

A fundamental idea in group theory, which ypu have enough tools to prove to yourself, is that the converse of this statement holds. A subgroup is normal if and only if it is the kernel of some homomorphism.

Further, I also claim that any homomorphism pushes down to an isomorphism. Consider a homomorphism from G to H. If K is the kernel of the homomorphism, then G/K is isomorphic to H.

Hopefully, this provides some motivation for the study of homomorphisms. In my first course on group theory, these results these results felt truly beautiful to me, I hope you feel the same.

Answer 3

A homomorphism is a map between two structures that preserves the operations defined on those structures. For instance, for simplicity, let $(G, \circ)$ and $(H, \diamond)$ be groups. Then a homomorphism is a mapping

$$\phi : G \to H$$

with the additional property that $\phi(g \circ h) = \phi(g) \diamond \phi(h)$ for $g, h \in G$. An isomorphism is a homomorphism which is bijective. This is useful because it means that the two structures are essentially the same. It also means we can move freely between the two structures because a bijective mapping has an inverse. That is, there is a map $\phi^{-1} : H \to G$ that is also a homomorphism. Two structures are isomorphic if there exists an isomorphism taking one to another.