inverse image of a morphism
InverseImageOfAMorphism
2008-09-02 01:01:48
2008-09-22 01:54:40
Definition
image of a morphism
inverse image
inverse coimage
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Let $f:A\to B$ be a morphism in a category $\mathcal{C}$. Let $\im(f)$ be the image of $f$ and $i:\im(f)\to B$ be a representing monomorphism. The inverse image of $f$ is the pullback of $f:A\to B$ and $i: \im(f) \to B$:
$$\xymatrix@+=3pc{
{C}\ar[r] \ar[d] &{A}\ar[d]^{f} \\
{\im(f)}\ar[r]^i &{B}
}
$$
$C$ is sometimes denoted by $f^{-1}(B)$. Since the diagram is a pullback and $i$ is monomoprhic, the inverse image $f^{-1}(B)$ is a subobject of $A$ (see \PMlinkname{this entry}{PullbackOfAMonomorphismIsAMonomorphism} for more detail.)
For example, in $\textbf{Set}$, the category of sets, the inverse image, in the sense above, of a morphism $f:A\to B$ is just the \PMlinkname{inverse image}{InverseImage} of $f$ as a function: clearly, $$f^{-1}(B)=\lbrace a\in A\mid f(a)\in B\rbrace$$ is a set (a subset of $A$). Let $j:f^{-1}(B)\to A$ be the canonical inclusion, and $\overline{f}: f^{-1}(B)\to \im(f)$ be the induced function by restricting the domain of $f$ to $f^{-1}(B)$ and the range to $\im(f)$. The diagram above is clearly commutative. Suppose there is a set $S$ and two functions $g:S\to A$ and $h:S\to \im(f)$ such that $f\circ g= i\circ h$. Define $k:S\to f^{-1}(B)$ by $k(s)=g(s)$. This is a well-defined function, since $f(g(s))=i(h(s))=h(s)\in B$, or $g(s)\in f^{-1}(B)$. Furthermore, $j(k(s))=j(g(s))=g(s)$, and $\overline(f)(k(s))=f(k(s))=f(g(s))=i(h(s))=h(s)$. Finally, it is easy to see that $k$ is unique.
\textbf{Remark}. The \emph{inverse coimage} of a morphism is dually defined.
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\bibitem{cf} C. Faith \emph{Algebra: Rings, Modules, and Categories I}, Springer-Verlag, New York (1973)
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