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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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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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