Skip to main content

One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the A1$A_1$ and B1$B_1$ from your diagram. If the maps from A2$A_2$ and A4$A_4$ to B2$B_2$ and B4$B_4$ are epimorphisms and the morphism A5 -> B5$A_5 \to B_5$ is monic then the cokernel of A3 -> A4$A_3 \to A_4$ is a subobject of the cokernel of B3 -> B4$B_3 \to B_4$. So morally B3$B_3$ is an "extension" of quotients of A2$A_2$ and A4$A_4$ and we have not "killed less stuff" in the bottom row so A3 -> B3$A_3 \to B_3$ should also be an epimorphism.

One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the A1 and B1 from your diagram. If the maps from A2 and A4 to B2 and B4 are epimorphisms and the morphism A5 -> B5 is monic then the cokernel of A3 -> A4 is a subobject of the cokernel of B3 -> B4. So morally B3 is an "extension" of quotients of A2 and A4 and we have not "killed less stuff" in the bottom row so A3 -> B3 should also be an epimorphism.

One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the $A_1$ and $B_1$ from your diagram. If the maps from $A_2$ and $A_4$ to $B_2$ and $B_4$ are epimorphisms and the morphism $A_5 \to B_5$ is monic then the cokernel of $A_3 \to A_4$ is a subobject of the cokernel of $B_3 \to B_4$. So morally $B_3$ is an "extension" of quotients of $A_2$ and $A_4$ and we have not "killed less stuff" in the bottom row so $A_3 \to B_3$ should also be an epimorphism.

Source Link
Greg Stevenson
  • 8.7k
  • 1
  • 40
  • 38

One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the A1 and B1 from your diagram. If the maps from A2 and A4 to B2 and B4 are epimorphisms and the morphism A5 -> B5 is monic then the cokernel of A3 -> A4 is a subobject of the cokernel of B3 -> B4. So morally B3 is an "extension" of quotients of A2 and A4 and we have not "killed less stuff" in the bottom row so A3 -> B3 should also be an epimorphism.