Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering what "a mapping factors over another mapping" generally means? Does it have something to do with commutative diagram in category theory?

I have seen this usage in different situations, and would like to give examples, but cannot recall them one by one except the most recent one from Fiber bundle:

Mappings which factor over the projection map are known as bundle maps.

Thanks and regards!

share|improve this question
1  
I would assume that $f$ factors over $g$ if there is some $h$ such that $f=h\circ g$ (or maybe $f=g\circ h$). I'm not certain, though. –  Asaf Karagila Aug 14 '11 at 13:11
    
Thie meaning is equivalent to the standard meaning in/with numbers: you have two maps f,g, and you want to find--if possible-- a "factorization" thru a map h so that fog=h. Depending on the context, there are many theorems telling you when this factoring is possible; if you are working with groups, then you need some condition of the kernels of certain maps; in other cases you may be working on a quotient space and you want a map to be constant in equivalent classes. –  gary Aug 14 '11 at 13:15
    
In the specific case you link to, bundle maps are maps from the base space B to the fiber, i.e., you choose b in B so that b is sent to its fiber $\pi^{-1}(b)$, and then you compose with the projection map $\pi$ to get b back. As an example, if your fiber is a vector space (as in a line bundle, aka vector bundle with dimension 1), a section maps b to the vector space associated with b under the projection; composing with the projection then, gives you b back. –  gary Aug 14 '11 at 13:24
    
To be more precise, you have two maps f,g both into a common space X, and you want to find a map h that "makes the diagram commute", i.e., you have maps between, say, f between A and X, g between B and X and you want to find a map h between A and B so that goh=f. This happens in many "categories", so I can only give a broad description --but , hey don't get me wrong, I love broads. –  gary Aug 14 '11 at 13:29
    
@gary: about the example in your broad description, for given f and g, can there be more than one mappings that make f factor over g? –  Tim Aug 14 '11 at 13:31

2 Answers 2

up vote 6 down vote accepted

As was pointed out in the comments:

Given a morphism $f:A \to C$ and a morphism $g: A \to B$ then $f$ is said to factor over $g$ if there exists $h: B \to C$ such that $hg = f$. Note that $h$ is unique as soon as $g$ is an epimorphism.

In the case of bundles $\pi_{E}: E \to M$ and $\pi_F: F \to N$ then a map $\varphi: E \to F$ is a bundle map if there exists $f:M \to N$ such that $\pi_{F} \varphi = \varphi \pi_{E}$, as is noted further down on the wikipedia page you linked to. So $\varphi : E \to F$ is a bundle map if and only if $\pi_{F} \varphi$ factors over $\pi_{E}$ via a (necessarily unique) map $f: M \to N$, as the bundle projections are assumed to be surjective. Note that this means in particular that $\varphi$ maps fibers to fibers.

Similarly, there is the notion of factoring through (you need to scroll down a little).

I'm mostly using this for the situation $f: A \to C$ and $h: B \to C$ then $f$ factors through $h$ if there is $g: A \to B$ such that $f = hg$. If $h$ happens to be a monomorphism then $g$ is unique (if it exists).

However, the distinction I'm making in this post are far from universally accepted.

You'll also find $f:A \to C$ factors through $B$ if there are maps $g: A \to B$ and $h: B \to C$ such that $f = hg$ and so on, $f$ factors through $g$, $f$ factors through $h$ in this same situation.

Basically, all it means is that $f$ can be decomposed as $f = hg$ in some way that should be obvious from the context. To repeat, if either $h$ is a monomorphism then $g$ is unique and if $g$ is an epimorphism then $h$ is unique.

share|improve this answer
    
There is also the abuse of terminology that I sometimes see (and use), where $f:A\to A$ and $h:A\to B$, and factor through is taken to mean $f$ "commutes" with $h$: $\exists g:B\to B$ so that $hf = gh$. Also, completely agree that usually the distinction between Theo's "through" and "over" is completely clear from context. –  Willie Wong Aug 14 '11 at 14:06
    
@Willie: right... That would be the special case of a bundle endomorphism, for example. What would you use for the situation $f: A \to A$ and $h: B \to A$ "$f$ lifts over $h$"? –  t.b. Aug 14 '11 at 14:09
    
I tend to think of $h$ as the map doing the lifting. So I would write "$h$ lifts $f$ to a map $g$" or "$f$ can be lifted by $h$ to a map $B\to B$". But that may be a personal quirk. –  Willie Wong Aug 14 '11 at 14:18

The most general, broader description of your question is that you have two sets $A,B$ and maps $f,g$ into a common third space $C$, i.e., you have ; $f:A\rightarrow C$, and you have $g\colon B\rightarrow C$, and then you want to see if there exists a map $h$ between $A$ and $C$, so that "the diagram commutes" (if you ever try to do algebraic topology you will see this everywhere; in analysis, almost everywhere ; ) , i.e., if you follow the arrows and do the composition from $A$ to $B$ to $C$ , composing $g\circ h$, you will get the same result as if you go along the arrow from $A$ to $C$ alone. It is difficult to generalize because the results will depend on the structure of $A,B$ and $C$ given.

In the case of bundles you mention, a bundle map associated with a bunlde $\pi\colon E\rightarrow B$ is a map that sends an element $b$ in the base to its fiber under $\pi$, so that, if you compose back the fiber element, you get $b$ back, i.e., $\pi \circ f(b)=b$ e.g., if your fiber is a vector space, your bundle map will send $b$ to an element in the vector space, so that the composition with $\pi$ will project back down to $b$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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