Show that if$f:A \to B$ is surjective and $h:A \to C$ such that $\ker(f) ⊆ \ker(h)$, then there exists $g:B \to C$ such that $h=g \circ f$. I was able to show if such a $g$ exists then it is unique and that $\ker(f) ⊆ \ker(h)$ (if such a $g$ exists however, I don't know why we need that $\ker(f) ⊆ \ker(h)$ for $g$ to exist. My homework asks us to show it is necessary and sufficient, I only see the one direction. (i.e. cant we just "define" g as such without worrying about its kernel?
Why can a map be factored out if $\ker(f) ⊆ \ker(h)$
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7What is the question? – 2012-09-10
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3This is called the kernel pair of $f$. It is worth noting that it defines an equivalence relation on $\operatorname{dom} f$, and the quotient of $\operatorname{dom} f$ by this equivalence relation has a natural bijection with $\operatorname{im} f$. It is in fact very interesting. – 2012-09-10
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0This isn't a "hard question", so I have some homework question about it but rather than posting that, I wanted to start a "discussion" about it, because this definition is very different than the one we are first introduced to in undergraduate group theory or linear algebra. – 2012-09-10
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4This website is not suitable for discussions – it is a question-and-answer site. – 2012-09-10
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0@ZhenLin In my class (and on wikipedia) they simply call it the kernel not a "kernel pair". Could you possibly demonstrate what you just said? – 2012-09-10
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0Right... then my "question" is. Can you demonstrate how it defines an equivalence relation and how there is a natural bijection,etc. – 2012-09-10
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0It _is_ an equivalence relation, if you think of binary relations on a set $X$ as being subsets of $X \times X$. The proof of this is easy and amounts to the fact that $=$ is an equivalence relation on the codomain. The fact that the quotient by this equivalence relation has a natural bijection with the image is more-or-less automatic once you realise the equivalence relation is "$x \sim y$ if and only if $f(x) = f(y)$". – 2012-09-10
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1@dustanalysis: if that *is* your question, maybe you should edit your question? – 2012-09-10
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6see also: [‘If your motivation for asking the question is “I would like to participate in a discussion about ______”, then you should not be asking here.’](http://math.stackexchange.com/faq#dontask) – 2012-09-10
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0This is no longer a "discussion" but a solid and clear question, I recommend you all open it or I will just post it again – 2012-09-10
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0@MichaelGreinecker: The question is on the first line. "Show that if $f:A \to B$ is surjective and $h : A \to C$ such that $\ker(f) ⊆ \ker(h)$, then there exists $g:B \to C$ such that $h=g \circ f$." – 2012-09-10
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0Yeah, I just edited it because (admittedly) I didn't want to ask this nor did I know that we weren't allowed to "discuss", this is my first post that sounds like a "discussion". Can some vote the question at least up to 0 because its a legit question now – 2012-09-10
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2There seems to be a question now; I’m voting to reopen. I don't see the point in asking it again separately, and in fact that would probably be discouraged. [Also, are these $A, B, C$ groups? Rings? It doesn't change much in the end, but it would be good to know.] – 2012-09-10
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0The question was for general sets, so I think the appropriate level might be "general algebras" (I don't think that, that imposes too much structure). – 2012-09-10
1 Answers
Suppose $\ker(f) ⊆ \ker(h)$. That is, $f(x)=f(y)$ implies $h(x)=h(y)$. For each $y\in B$, the function $h$ is constant on the set $\{x\in A:f(x)=y\}$. So we can define a function $g:B\to C$ that maps $y\in B$ to the unique point $c\in C$ that every element of $\{x\in A:f(x)=y\}$ gets mapped to. By construction, $h=g\circ f$.
Now for each $y=f(x)$ the equation $h(x)=g(f(x))$ implies that $g$ is uniquely defined at a point of the form $f(x)$, that is the value is uniquely determined at the range of $f$. Since $f$ is surjective, every point is of this form. Otherwise, one could define $g$ arbitrarily at points in $B\backslash f(X)$
Now suppose there exists $g:B\to C$ such that $h=g\circ f$. Then, whenever $f(x)=f(y)$, we have $g(f(x))=g(f(y))$ which is equivalent to $h(x)=h(y)$ and hence $\ker(f) ⊆ \ker(h)$.
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0If we remove the fact that f is surjective, could it be the case that ker(h) is in ker(f)? In other words, for the top part of the proof I still dont see why we need it to be surjective (I know why we need it in the other direction) – 2012-09-10
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0@dustanalysis Yes, the only thing that changes is that there could be more than one function $g$ such that $h=g\circ f$. The existence of such a function is still equivalent with $\ker(f) ⊆ \ker(h)$. – 2012-09-10
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0OKay.... I think I've got it. If f were not surjective, then h may not be "well-defined" (it may not "see" all of B after composition for some f(x)). In particular, ker(f) in ker(h) may be a nonsensical statement. – 2012-09-10
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1@dustanalysis The statement $\ker(f) ⊆ \ker(h)$ makes still sense if $f$ is not surjective. Let $A=B=C=\{0,1\}$. The function $f$ maps both elements to $0$ and the function $h$ maps both elements to $1$. Now we have $\ker(f) ⊆ \ker(h)$, but $f$ is not surjective, so there exists more than one function $g$ such that $g\circ f=h$. One possibility is that $g$ maps, like $h$, both elements to $1$. The other is that $g$ maps $1$ to $0$ and $0$ to $1$. – 2012-09-10