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image of a morphism
Image
Let $\mathcal{C}$ be a category and a $f:A\to B$ a morphism from objects $A$ to $B$ in $\mathcal{C}$. An image of $f$ is a subobject $I$ of $B$ with a representing monomorphism $g:I\to B$, such that

$f$ factors through $g$; i.e. there is a morphism $h:A\to I$ such that $f=g\circ h$:
$A\buildrel f\over{\longrightarrow}B=A\buildrel h\over{\longrightarrow}I% \buildrel g\over{\longrightarrow}B$ 
if $f$ factors through a monomorphism $s:J\to B$:
$A\buildrel f\over{\longrightarrow}B=A\buildrel t\over{\longrightarrow}J% \buildrel s\over{\longrightarrow}B,$ then there is a morphism $x:I\to J$ such that
$I\buildrel g\over{\longrightarrow}B=I\buildrel x\over{\longrightarrow}J% \buildrel s\over{\longrightarrow}B.$
Example. In the category of sets, the image of a function $f:A\to B$ is just the image $f(A)$ with the canonical injection into $B$.
In the literature, the first bullet is equivalent to saying that the subobject $g:C\to B$ allows $f$. So the definition of the image of $f$ is the smallest subobject of $B$ that allows $f:A\to B$.
Coimage
Dually, one defines the coimage of $f:A\to B$ as a quotient object $C$ of $A$ with a representing epimorphism $h:A\to C$, such that

$f$ factors through $h$; i.e. there is a morphism $g:C\to B$ such that $f=g\circ h$:
$A\buildrel f\over{\longrightarrow}B=A\buildrel h\over{\longrightarrow}C% \buildrel g\over{\longrightarrow}B$ 
if $f$ factors through an epimorphism $t:A\to D$:
$A\buildrel f\over{\longrightarrow}B=A\buildrel t\over{\longrightarrow}D% \buildrel s\over{\longrightarrow}B,$ then there is a morphism $y:D\to C$ such that
$A\buildrel h\over{\longrightarrow}C=A\buildrel t\over{\longrightarrow}D% \buildrel y\over{\longrightarrow}C.$
Remarks.
1. In the definition of image, since $g$ is a monomorphism, $x$ is a monomorphism. Furthermore, since $s$ is a monomorphism, $x$ is uniquely determined, and we have the following commutative diagram:
$\xymatrix@R=2pt{&I\ar[dd]^{x}\ar[dr]^{g}&\\ A\ar[ur]^{h}\ar[dr]_{t}&&B\\ &J\ar[ur]_{s}&}$
2. Dually, $y$ is a uniquely determined epimorphism satisfying the following commutative diagram:
$\xymatrix@R=2pt{&D\ar[dd]^{y}\ar[dr]^{g}&\\ A\ar[ur]^{t}\ar[dr]_{h}&&B\\ &C\ar[ur]_{s}&}$
3. If $f:A\to B$ has an image (dually, coimage), it is unique up to isomorphism. The image and coimage of $f$ are denoted by $\operatorname{im}(f)$ and $\operatorname{coim}(f)$ respectively.
4. Suppose that a category is Ab1. If $\operatorname{im}(f)$ and $\operatorname{coim}(f)$ exist for $f:A\to B$, then there is a unique morphism $\overline{f}:\operatorname{coim}(f)\to\operatorname{im}(f)$ such that
$\xymatrix@C+=4pc{{A}\ar[r]^{{f}}\ar[d]&{B}\\ {\operatorname{coim}(f)}\ar[r]^{{\overline{f}}}&{\operatorname{im}(f)}\ar[u]}$ is commutative. The Ab2 Axiom, a la Grothendieck, is the statement that if $\overline{f}$ exists, it is an isomorphism. A category is said to be Ab2 if it is Ab1, and every morphism satisfies the Ab2 Axiom.
5. A category $\mathcal{C}$ is said to have images (dually, has coimages) if the image (coimage) of any morphism exists.
Every abelian category has images and coimages, and
$\operatorname{im}(f)=\ker(\operatorname{cok}(f))\qquad\mbox{ and }\qquad% \operatorname{coim}(f)=\operatorname{cok}(\ker(f)),$ 
where $\ker$ and $\operatorname{cok}$ are the kernel and cokernel operations. In addition, we have the following important result:
if a morphism $f:A\to B$ can be factored as
with $g$ a monomorphism and $h$ an epimorphism, then $I$ (with $g$) is the image of $f$ and $I$ (with $h$) is the coimage of $f$.
$A\buildrel f\over{\longrightarrow}B=A\buildrel h\over{\longrightarrow}I% \buildrel g\over{\longrightarrow}B$
In other words, the factorization above is uniquely determined, up to isomorphism.
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