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Comutative Diagram composed by exact sequences and R-mods

I got an exercise from my introduction to Modules class, so the image is a commutative diagram where each row forms an exact sequence and every object is an $R-$mod, the question is to prove that if $h_2$ is surjective and $h_3$ is injective then $Im(h_4)=g_3(Im(h_3))$. Without prior experience with chasing diagrams around except the previous exercises in a similar vain, it seems this is false and the question is wrong.
I failed to find a counter-example trying with some geometric linear morphisms in Euclidean spaces but my mind got really out of it after some failed attempts.
I know this is very closely related to the Five-Lemmas and Diagram chasing in general but I didn't find the answer after a quick look online, none of the results I found involved going back without any information of the functions afterwards.

David Melo
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  • If we take $M_2=M_3=N_3=N_4$ and all other modules to be the zero module (with the obvious maps), it seems to provide a counterexample. – Greg Martin Oct 03 '20 at 19:07
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    That fails commutativity on the second and third square, I'll keep trying. – David Melo Oct 03 '20 at 19:19
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    It looks false. A nevessary condition is that $g_4 \circ h_4=0$, and I can’t see how the assumptions imply that, eg take $N_i=M_i$. – Aphelli Oct 03 '20 at 19:24
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    Indeed, @Mindlack's comment shows that in the situation where all the $h_j$ are isomorphisms, the statement is claiming that $g_3$ is surjective always, under no hypotheses other than (horizontal) exactness. So a counterexample then (my second attempt...) is $M_4=M_5=N_4=N_5$ and all other modules zero, with the obvious maps. – Greg Martin Oct 03 '20 at 20:02

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