27
$\begingroup$

I'm starting to learn category theory, but there's one thing I don't get: all morphisms seem to be homomorphisms; the definition seems to be the same. What's the difference between these two? Can you give me an example of a non-homomorphic morphism?

Thank you for your patience,

$\endgroup$
  • 6
    $\begingroup$ Stating the definitions seems appropiate here. $\endgroup$ – Git Gud Jul 7 '13 at 19:22
  • 3
    $\begingroup$ I think this is a bit like asking for an example of a group which doesn't have multiplication as its binary operation. If you're going to use a term like 'homomorphism' outside of its usual realm of usage (ie functions which preserve algebraic structures of objects) then you need to define exactly what you mean by 'homormophism' and 'structure preserving'. $\endgroup$ – Dan Rust Jul 7 '13 at 19:44
  • 2
    $\begingroup$ What do you mean by 'structure'? By mapping do you mean function? Because there are many good examples of categories where the arrows are not functions. $\endgroup$ – Dan Rust Jul 7 '13 at 19:56
  • 3
    $\begingroup$ In some categories, objects don't even have elements (posets for example or a suitable discrete category). So, it really is better to not think about objects having elements at all. $\endgroup$ – Dan Rust Jul 7 '13 at 20:20
  • 3
    $\begingroup$ It depends what the other morphisms are and how composition is defined. The whole point is that a morphism isn't defined by how it acts on elements, but by how it composes with other morphisms (in fact the former can be defined in terms of the latter). $\endgroup$ – Dan Rust Jul 7 '13 at 21:01
26
$\begingroup$

First, keep in mind that the term "homomorphism" predates both the term "morphism" and the creation of category theory.

"Homomorphism," roughly speaking, refers to a map between sets equipped with some kind of structure that preserves that structure. The collection of all such structured sets and homomorphisms between them thus forms a category $C$, but this category is itself equipped with some extra structure, namely a faithful functor $C \to \text{Set}$, making it a concrete category.

"Morphism," on the other hand, refers to an arrow in an arbitrary category. In particular, an arbitrary category need not be concretized (and in fact some categories cannot be concretized!). Its objects need not be interpreted as sets and its morphisms need not be interpreted as functions: there are many other ways of writing down categories than thinking about structured sets.

Here is an example of a "category where the morphisms are not homomorphisms," but you really shouldn't think of it that way: consider the category whose objects are groups and whose morphisms are all functions (not necessarily preserving the group operations). (The reason I say you shouldn't think of it that way is that, to a category theorist, objects are just placeholders describing how morphisms can compose: all of the important information is in the morphisms, so I have no right to call the objects in this category groups if I can't detect their group structure using morphisms.)

Concretizations of categories are both analogous to and generalize group actions. One of the more important habits of thought you can pick up while learning category theory is to think about categories abstractly without choosing a concretization, in the same way that you think about groups abstractly without choosing a group action.

$\endgroup$
  • 1
    $\begingroup$ So homomorphism is like half-way the generalization between a function and a morphism? (BTW, I indeed have heard, actually read, of those concrete categories) $\endgroup$ – JMCF125 Jul 7 '13 at 20:34
  • 1
    $\begingroup$ A function is a structure-preserving homomorphism. It preserves the structure of being a set (if you think of the groups as being built from sets), or more generally the equality relation between elements of the group. As you noted, arbitrary functions are not the right notion of morphism in the category of groups, if the latter is intended to capture the usual content of group theory. But they do not provide a distinction between morphism and homomorphism. $\endgroup$ – zyx Dec 20 '15 at 0:39
19
$\begingroup$

There are already some excellent answers here, but let me add something from a slightly different point-of-view.

First of all, homomorphism is not a word that can be used nakedly (so to speak) in mathematics. It is a function between two structures that preserves their structure, in some sense, and this sense has to be specified as part of the theory of those particular structures. (E.g. in the theory of groups, we define group homomorphisms; in the theory of rings, we define ring homomorphisms, etc.)

Also, in certain contexts, there can be competing notions of homomorphisms (for rings with identity, should the homomorphisms preserve the identity?; for Banach spaces, should morphisms be simply continuous, or should they preserve the norm?); in any particular situation, if it hasn't already been made clear in the context, you have to specify which definition you're using.

Now, in some situations, we don't use the word homomorphism; e.g. we don't speak of a homomorphism of topological spaces (instead, we have continuous maps between topological spaces). The choice of whether or not to use the word homomorphism is to some extent dictated by tradition, and also by general mathematical culture; the more algebraic the structures under consideration are, the more likely the word homomorphism is to be used in regard to them.

Now category theory abstracts the notion of mathematical structure and structure preserving maps between them. As already noted in other answers, not all categories are concrete, or, even if they are, are not naturally thought of as being concrete. And if we're talking about an arbitrary category, we don't specify what sort of objects the objects actually are; we just write $\mathcal C$, and $x \in\operatorname{Ob}\, \mathcal C$, and $f: x \to y$, without specifying what $\mathcal C$ is (just as in the theorems of a group theory text, we don't specify what the group $G$ is or what it's elements $x$ are or what its operation is; it is just some abstract, unspecified group).

Now the founders of category theory knew that they were abstracting a situation in which very often the category was going to consist of algebraic structures and homomorphisms between them. But they also knew that they were abstracting examples like topological spaces and continuous maps, where homomorphism is not the traditional term. Hence, they chose morphism as being close enough to homomorphism to be suggestive, but not identical, so that it didn't give the psychological impression of ruling out contexts like topological spaces and continuous maps. [This last paragraph is my invention, not actual history, but is probably fairly close to the actual history, and hopefully gets the point across.]

Finally, it seems worth noting that often now in mathematics people speak of a morphism of rings or groups or ..., i.e. they drop the prefix homo-, and I think this is the influence of the general category-theoretic terminology.

$\endgroup$
12
$\begingroup$

I will proceed to specify a category $\mathcal{C}$:

  • The objects of $\mathcal{C}$ are as follows: $\;\;\mathrm{ob}(\mathcal{C})=\{\star,\diamond\}$

  • The morphisms of $\mathcal{C}$ are as follows:

    $$\begin{align*} \mathrm{Mor}_{\mathcal{C}}(\star,\star)&=\{\mathrm{id}_{\star}\}& \mathrm{Mor}_{\mathcal{C}}(\diamond,\diamond)&=\{\mathrm{id}_{\diamond}\}\\[0.1in] \mathrm{Mor}_{\mathcal{C}}(\star,\diamond)&=\{\heartsuit\}&\mathrm{Mor}_{\mathcal{C}}(\diamond,\star)&=\{\bullet\} \end{align*}$$

Are the morphisms of $\mathcal{C}$ homomorphisms?

(What I'm getting at is: what would it even mean to be a homomorphism from $\star$ to $\diamond$?)

$\endgroup$
  • 5
    $\begingroup$ I mean that categories can consist of objects which have no internal structure, and therefore have no corresponding concept of "homomorphisms". $\endgroup$ – Zev Chonoles Jul 7 '13 at 19:32
  • 1
    $\begingroup$ What do you mean by "transform"? The entirety of information contained in (for example) the morphism $\heartsuit$ is: $$\begin{array}{l} \sf\bullet\;\text{ its name}\\ \sf\bullet\;\text{ which object it starts at}\\ \sf\bullet\;\text{ which object it ends at}\\ \end{array}$$ There is no "transforming" going on. The symbol $\heartsuit$ does not (necessarily) denote a function because $\star$ and $\diamond$ don't (necessarily) denote sets. $\endgroup$ – Zev Chonoles Jul 7 '13 at 19:42
  • 2
    $\begingroup$ Nope - I just like that particular shade of blue :) $\endgroup$ – Zev Chonoles Jul 7 '13 at 20:35
  • 2
    $\begingroup$ Guess what my computer wallpaper is? :) $\endgroup$ – Zev Chonoles Jul 7 '13 at 20:42
  • 3
    $\begingroup$ @JMCF125 Because in mathematics, when you use a term, you have to define what it means! You are free to come up with a definition of "relation" between anything, but you have to have a definition. $\endgroup$ – us2012 Jul 7 '13 at 21:42
7
$\begingroup$

The morphisms in a category do not have to be functions at all, let alone homomorphisms. All that is required of a morphism in a category is to know what its domain and codomain are, and how to compose it with all morphisms in the category.

Examples of categories where morphisms are not functions: Any poset $(P,\le)$ can be considered as a category where the objects are the elements in $P$ and a morphisms $x\to y$ exists iff $x\le y$.

The category $Rel$, whose objects are all sets and a morphisms $f:A\to B$ is a relation, i.e. a subset $f\subseteq A\times B$.

The category $Mat$ whose objects are the natural numbers, and a morphism $n\to m$ is an $n\times m$ matrix (with real entries for simplicity). Composition is matrix multiplication.

$\endgroup$
  • $\begingroup$ Homomorphisms have to be functions? That's not what I read. $\endgroup$ – JMCF125 Jul 7 '13 at 19:28
  • 2
    $\begingroup$ @JMCF125 what then are homomorphisms? $\endgroup$ – Ittay Weiss Jul 7 '13 at 19:30
  • $\begingroup$ They're mappings between mathematical structures that preserve their structure, as I wrote in a comment above. That's why I don't understand what differs them from morphisms. $\endgroup$ – JMCF125 Jul 7 '13 at 19:37
  • 7
    $\begingroup$ mappings are typically taken to be synonymous with functions. Anyways, you should really define your terms then more clearly. Besides, you were already given several examples where morphisms in a category are not structure preserving anything. $\endgroup$ – Ittay Weiss Jul 7 '13 at 19:40
2
$\begingroup$

The term 'homomorphism' usually refers to structure-preserving maps between algebraic objects like groups, rings, and vector spaces. Hence, in categories like $\mathbf{Grp}$, $\mathbf{Ring}$, and $\mathbf{Vect}_k$, the morphisms can be called homomorphisms.

But in other categories like $\mathbf{Top}$, the category of all topological spaces, the morphisms are continuous functions, which aren't really referred to as homomorphisms. In fact, a morphism doesn't need to be a function at all. Fix a commutative ring $R$. The category $\mathbf{Matr}_R$ consists of all positive integers as objects, and each $m \times n$ matrix $A$ with entries from $R$ is regarded as a morphism $A : n \to m$.

So I think you're confused. The term 'homomorphism', whenever applicable, means the exact same thing as morphism.

$\endgroup$
1
$\begingroup$

Not so. The definition of a "morphism" depends heavily on the category in question - in fact, it is part of the definition of the category.

So, for instance, you could consider the category $SET$, where the objects are sets and the morphisms between sets $A$ and $B$ are simply all the functions possible between the two sets.

$\endgroup$
  • 1
    $\begingroup$ but functions between sets are 'homomorphisms'. It's just that there is no structure to preserve, so all functions preserve structure. $\endgroup$ – Ittay Weiss Jul 7 '13 at 19:28
  • $\begingroup$ I don't mean in a specific category. I mean, generally, isn't there a difference between those two? Kind of in the way there is a different definition of epimorphism and monomorphism. $\endgroup$ – JMCF125 Jul 7 '13 at 19:29
  • 2
    $\begingroup$ You could certainly look at it that way. But you could also define a class where the objects are finite abelian groups and the morphisms are still simply all functions between them as sets. You won't see things like this terribly often, because morphisms that preserve structure tend to be more useful. However, all that is required of morphisms is that there be an identity morphism for each object and that morphism composition be associative. $\endgroup$ – Nick Peterson Jul 7 '13 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.