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I need to prove:

Let $C$ be a category with equalizer , coequalizer, kernels, cokernels, normal and conormal. Let $f : A \to B$ be an epimorphism and $u : B' \to B$ a subobject of $B$. Then $f(f^{-1}(B')) = B'$.

My attempt:


\begin{array}{ccc} & & B' \\ & & \downarrow u \\ A & \xrightarrow{f} & B \end{array}

Since $\mathcal{C}$ is normal and has kernels, it has inverse images. Let us consider:


\begin{array}{ccc} f^{-1}(B') & \to & B' \\ \downarrow & & \downarrow u \\ A & \xrightarrow{f} & B \end{array}

Then $f^{-1}(B') \cong$ subobject of $A$ and is the largest object mapped to $B$ via $f$.

Since $\mathcal{C}$ is conormal, $f: A \to B$ being an epimorphism is the cokernel of some morphism $f': A' \to A$. Therefore, we have the sequence

$A' \xrightarrow{f'} A \xrightarrow{f} B \to B$

Then we have that $f f'$ is the kernel of $g$, since $f f' = 0$, hence $g f f' = 0$. I don't know how can I follow, why I need equalizers and coequalizer? Any hint?

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  • $\begingroup$ I assume your category is preadditive? Or is it even abelian? $\endgroup$
    – Martin Brandenburg
    Commented Nov 19 at 16:11
  • $\begingroup$ @MartinBrandenburg my teacher only said me that I need to prove the category is exact $\endgroup$
    – C25
    Commented Nov 20 at 23:06
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    $\begingroup$ It seems to me that you would need to assume that cokernels or regular epimorphisms are pullback-stable to get your results, is that what you mean by $\mathcal{C}$ being normal? $\endgroup$
    – Arnaud D.
    Commented Nov 21 at 14:26
  • $\begingroup$ Yes, that's normal. $\endgroup$
    – C25
    Commented Nov 27 at 17:57

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