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Given a function $f:X\to Y$ in category $\mathcal{C}$, one can construct the image as a factorisation $f=(e:I\hookrightarrow Y)\circ(g:X\to I)$ that is universal (initial) among all such factorisations.

This does seem like a universal property. But I can't figure out how this can actually be constructed as an initial object in a comma category, because there are morphisms both from and to the object.

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  • $\begingroup$ It is not guaranteed that each morphism has an image. $\endgroup$ – Paul Frost Jan 23 at 13:05
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Let $\newcommand\Sub{\operatorname{Sub}}\Sub(Y)$ the preorder of subobjects of $Y$ in $\mathcal C$. Then the "image functor" can be seen as a left-adjoint of the forgetful functor $\mathcal C/Y\leftarrow\Sub(Y)$.

In particular, if $f=e\circ g$ is an image factorization of $f$, then $g:(X,f)\to(I,e)$ is a universal arrow respect to the forgetful functor $\mathcal C/Y\leftarrow\Sub(Y)$.

If $\mathcal C/Y\leftarrow\Sub(Y):\Gamma$ denote such forgetful functor, then $((X,f),g)$ is an initial object in the comma category $(X,f)\downarrow\Gamma$.

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