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 this entry for more detail.)

For example, in ${Set}$ the category of sets, the inverse image, in the sense above, of a morphism $f:A\to B$ is just the inverse image 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.

Remark. The inverse coimage of a morphism is dually defined.

Bibliography

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C. Faith Algebra: Rings, Modules, and Categories I, Springer-Verlag, New York (1973)