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# inverse image of a morphism

Let $f:A\to B$ be a morphism in a category $\mathcal{C}$. Let $\operatorname{im}(f)$ be the image of $f$ and $i:\operatorname{im}(f)\to B$ be a representing monomorphism. The inverse image of $f$ is the pullback of $f:A\to B$ and $i:\operatorname{im}(f)\to B$:

$\xymatrix@+=3pc{{C}\ar[r]\ar[d]&{A}\ar[d]^{{f}}\\ {\operatorname{im}(f)}\ar[r]^{i}&{B}}$ |

$C$ 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,

$f^{{-1}}(B)=\{a\in A\mid f(a)\in B\}$ |

is a set (a subset of $A$). Let $j:f^{{-1}}(B)\to A$ be the canonical inclusion, and $\overline{f}:f^{{-1}}(B)\to\operatorname{im}(f)$ be the induced function by restricting the domain of $f$ to $f^{{-1}}(B)$ and the range to $\operatorname{im}(f)$. The diagram above is clearly commutative. Suppose there is a set $S$ and two functions $g:S\to A$ and $h:S\to\operatorname{im}(f)$ 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.

# References

- 1
C. Faith
*Algebra: Rings, Modules, and Categories I*, Springer-Verlag, New York (1973)

## Mathematics Subject Classification

18A05*no label found*

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