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Suppose we're in the category $\mathbf{Set}$. Then the kernel of a pair of morphisms $f,g\colon A\to B$ is a third morphisms $h\colon C\to A$ such that for morphism $k\colon D\to B$ such that $fk=gk$, there is a unique $j\colon D\to C$ such that $hj=k$.

What exactly is $h\colon C \to A$ in this category? I'm curious what the set $C$ and function $h$ would be, and then I think I could verify that it actually satisfies the desired properties.

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What are your thoughts? If $f\circ k=g\circ k$, then what can you say about the image of $k$? –  Henning Makholm Nov 16 '13 at 21:10
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@HenningMakholm Then $f$ and $g$ agree on the image of $k$. I was thinking something along the lines of defining $C=\{a\in A:f(a)=g(a)\}$, and letting $h$ be the imbedding of $C$ in $A$. –  Nastassja Nov 16 '13 at 21:11
    
x @Natassja: Yes, that will work. (And you beat the first answer by just 5 seconds :-) –  Henning Makholm Nov 16 '13 at 21:12
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This is more often called "equalizer" rather than kernel –  Najib Idrissi Nov 16 '13 at 21:20
    
Thanks @Henning, nik. –  Nastassja Nov 16 '13 at 21:22

1 Answer 1

up vote 3 down vote accepted

$h$ is essentially the inclusion of $C=\{x\in A:f(x)=g(x)\}$; i.e. exactly the set of inputs on which $f$ and $g$ agree.

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Thanks, Malice. –  Nastassja Nov 16 '13 at 21:22

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