Let $\mathcal C$ be a category with images à la this Wiki article.

Let $f \colon A \longrightarrow B$ and $g: B\longrightarrow C$ be two morphisms. Let $X \xrightarrow{r} A$ be a subobject. Then $f(X)$ is defined to be the image of $fr$. Is it true that $$ (gf)(X) = g (f(X))?$$ I can show that $(gf)(X) \subseteq g(f(X))$ since it is reasonably easy to draw an arrow using the universal property of the image.

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    $\begingroup$ I don't think it is true in general, but it is true for categories which are balanced and have epimorphic images [Mitchell, Theory of Categories] $\endgroup$ – Paul Slevin Dec 19 '12 at 23:48
  • $\begingroup$ "epimorphic images" - How do you mean that? $\endgroup$ – Berci Dec 20 '12 at 1:31
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    $\begingroup$ Images are in general not useful without some structure on the ambient category. If $\mathcal{C}$ has pullbacks, then images behave sufficiently well that they compose as you expect. $\endgroup$ – Zhen Lin Dec 20 '12 at 2:02
  • $\begingroup$ @Berci I misspoke. I meant to say that the category has epi-mono factorisations $\endgroup$ – Paul Slevin Dec 20 '12 at 8:54
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    $\begingroup$ If $\mathcal{C}$ has pullbacks, then images are functorial, in the sense that pulling back along a morphism $f : A \to B$ produces a functor $f^{-1} : \textrm{Sub}(B) \to \textrm{Sub}(A)$, and then the left adjoint $\exists_f : \textrm{Sub}(A) \to \textrm{Sub}(B)$ (if it exists) gives images. Of course, left adjoints are unique up to unique isomorphism if they exist, so $\exists_g \exists_f \cong \exists_{g \circ f}$. $\endgroup$ – Zhen Lin Dec 20 '12 at 16:35

As I find myself often saying on this site, the notion of ‘image’ in a category without enough structure is often quite badly behaved and not interesting.

First, a positive answer. Let $\mathcal{C}$ be a category with pullbacks (or at least, pullbacks of monomorphisms). Then for each morphism $f : A \to B$ in $\mathcal{C}$ we get a pullback functor $f^{-1} : \textrm{Sub}_\mathcal{C}(B) \to \textrm{Sub}_\mathcal{C}(A)$. If $\mathcal{C}$ has image factorisations, then every such functor has a left adjoint, $\exists_f : \textrm{Sub}_\mathcal{C}(A) \to \textrm{Sub}_\mathcal{C}(B)$ that sends a subobject $A' \rightarrowtail A$ to the image of $A' \rightarrowtail A \xrightarrow{f} B$. Now, it is well-understood fact that left adjoints are unique up to unique isomorphism, and the pullback pasting lemma implies $f^{-1} g^{-1} \cong (g \circ f)^{-1}$, therefore $\exists_g \exists_f \cong \exists_{g \circ f}$, as expected. In particular, this is true when $\mathcal{C}$ is a regular category, which is the usual setting for studying image factorisations.

Now, a negative answer. Consider the following category:

  • Objects are $T, A, B, C, B', C', C''$.
  • Morphisms are $x, y, z : T \to A$, $f : A \to B$, $g : B \to C$, $\overline{f} : A \to B'$, $b : B' \to B$, $c : C' \to C$, $c' : C'' \to C'$, $\overline{g \circ b} : B' \to C'$, $\overline{g \circ f} : A \to C''$, plus their composites, subject to these equations:
    • $f = b \circ \overline{f}$
    • $g \circ b = c \circ \overline{g \circ b}$
    • $\overline{g \circ b} \circ \overline{f} = c' \circ \overline{g \circ f}$
    • $\overline{f} \circ x = \overline{f} \circ y$
    • $\overline{g \circ f} \circ x = \overline{g \circ f} \circ y = \overline{g \circ f} \circ z$

By realising $\mathcal{C}$ as a certain non-full subcategory of $\textbf{Set}$ (say, $T = \{0\}$, $A = \{ 1, 2, 3 \}$, $B' = \{ 4, 5 \}$, $B = \{ 6, 7 \}$, $C'' = \{ 8 \}$, $C' = \{ 9 \}$, $C = \{ 10 \}$), one can verify the following inequations:

  • $x \ne y$, $y \ne z$, $z \ne x$.
  • $\overline{f} \circ y \ne \overline{f} \circ z$.

Thus, $f, \overline{f}, g, g \circ f, \overline{g \circ f}, g \circ b, \overline{g \circ b}, \overline{g \circ b} \circ \overline{f}$ are not monic, while the rest are. It is not hard to check that we have the following image factorisations: \begin{align} f & = b \circ \overline{f} & \overline{f} & = \textrm{id}_{B'} \circ \overline{f} & g & = \textrm{id}_C \circ g \\ g \circ f & = (c \circ c') \circ \overline{g \circ f} & \overline{g \circ f} & = \textrm{id}_{C''} \circ \overline{g \circ f} \\ g \circ b & = c \circ \overline{g \circ b} & \overline{g \circ b} & = \textrm{id}_{C'} \circ \overline{g \circ b} & \overline{g \circ b} \circ \overline{f} & = c' \circ \overline{g \circ f} \\ \end{align} In particular, the image of $g \circ f$ is not isomorphic to the image of $g \circ b$, even though $b$ is the image of $f$. Unsurprisingly, this must be because the category doesn't have enough pullbacks, and sure enough, the pullback of $\overline{g \circ b}$ along $c'$ does not exist.

  • $\begingroup$ Good to see a counterexample. Nice answer, Zhen! $\endgroup$ – Michal R. Przybylek Jan 4 '13 at 23:23

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