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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!

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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
up vote 9 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.

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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 $b$.

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