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Consider the following diagram:

commutative diagram

What does it mean precisely to say "$f$ factors through $G/\text{ker}(f)$"?

Does it mean $f = \tilde{f} \circ \pi$, for some $\tilde{f}$?

I've seen texts use the phrase, but never a definition of this notion.

1 Answers 1

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It means exactly what you write: that you can express $f$ as "product" (composition) of two functions, with the first function going through $G/\mathrm{ker}(f)$; by implication, that map will be the "natural" map into the quotient, i.e., $\pi$. Under more general circumstances, you would also indicate the map in question.

The reason for the term "factors" is that if you write composition of functions by juxtaposition, which is fairly common, then the equation looks exactly as if you "factored" $f$: $f=\tilde{f}\pi$.

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    In general, how can we be assured that this happens? For instance, if we replace $G/\text{ker}(f)$ with a subgroup $H < G$, what are the requirements on $H$ (or $\pi$?) to ensure $f$ factors through $H$?2011-02-14
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    You are already using $H$ in the commutative diagram, so I assume you mean some random subgroup $K$. There is no canonical homomorphism from $G$ to a subgroup, so this is a situation where you would *need* to indicate what map $g\colon G\to K$ explicitly. In order to be able factor through such a $g$, the kernel of $f$ must contain the kernel of $g$. But this is not sufficient, because it may be impossible to map $K$ to $H$ (this may happen if your $g$ is not onto). If $g$ is onto and the kernel of $g$ contained in the kernel of $f$, then you can factor; $K$ need not be a subgroup.2011-02-14
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    A morphism $f\colon G\to H$ **always** factors through $G/\mathrm{ker}(f)$. This is sometimes called the First Fundamental Theorem of Homomorphisms.2011-02-14