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I'm reading "A survey of Semi-abelian Categories" of F. Borceux and I'm struggling with a proof for hours now. I do not think specific skills in semi-abelian categories are required to help me here, just basic skills on pullbacks.enter image description here

The part I do not understand is "When the right hand square is a pullback, so is the left hand square". I can not prove it ... In another reference they use a metatheorem by just proving it in $\mathbf{Set}$ but I want to write it only with morphisms, in a categorical way. Can someone help me on this one ?

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Sov
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    Kernel pairs are obtained as pullbacks; using this, it's just a matter of applying the "pullback pasting theorem" to the right diagram (try to draw a cube). – Arnaud D. Oct 24 '18 at 12:42
  • The cube I alluded to above is a bit like the one mentionned at https://math.stackexchange.com/questions/2829677/pullback-of-a-pullback-square-along-f-is-again-a-pullback-square. – Arnaud D. Oct 24 '18 at 12:47

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$\require{AMScd}$ In the following diagram, both squares are pullback by assumption, hence, by pullback pasting theorem, the outer rectangle is a pullback as well: \begin{CD} K[f']@>p'_i>>A'@>\beta>> A\\ @Vp'_iVV @VVf'V @VVfV\\ A'@>>f'>X'@>>\alpha>X \end{CD} In the following diagram, the outer rectangle (which is the same as before) and the rigth-hand square are pullbacks, hence, by pullback pasting theorem, the left-hended square is a pullback as well. \begin{CD} K[f']@>\gamma>>K[f]@>p_i>> A\\ @Vp'_iVV @VVp_iV @VVfV\\ A'@>>\beta>A@>>f>X \end{CD}

Fabio Lucchini
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  • I can not think I spent so much time for this... I should be able to find it myself. Thank you for the answer. – Sov Oct 24 '18 at 13:05