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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.

Andrés E. Caicedo
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Corey Harris
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1 Answers1

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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" $f$: $f=\tilde{f}\pi$.

Arturo Magidin
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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 $H$? – Corey Harris Feb 14 '11 at 13:45
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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. – Arturo Magidin Feb 14 '11 at 14:13
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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. – Arturo Magidin Feb 14 '11 at 14:16
  • what is the equivalent of factors through $K$ for the post-composition case, namely, when $K$ is on the left hand of the composition? – John Mars Mar 10 '21 at 15:58
  • @JohnMars I’m sorry, but I don’t understand what you are asking. – Arturo Magidin Mar 10 '21 at 16:44
  • When we say $f$ factors through $k$ we mean, as you say well, that there exists $\bar{f}$ such that $f = \bar{f}k$. There, k is on the right-hand side of the composition ($ \bar{f}k$). I'm working with something analogous, but k is on the left-hand side. Because of it, I'm looking for a linguistic expression for the left-hand case. – John Mars Mar 10 '21 at 17:15
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    @JohnMars: It is the exact same thing. We also say $f$ factors through $k$. Note that in general only one of the compositions will make sense (since codomain must match domain), so the expression will not be ambiguous most of the time. – Arturo Magidin Mar 10 '21 at 17:20
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    @JohnMars: for example, the universal property of the product $G\times H$ is that given any group $M$ and morphisms $g\colon M\to G$ and $h\colon M\to H$, there exists a unique map $p\colon M\to G\times H$ such that $g$ factors through $\pi_G$ ($g=\pi_G p$) and $h$ factors through $\pi_H$ ($h=\pi_H p$). – Arturo Magidin Mar 10 '21 at 17:26