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Consider chain of linear maps between finitely dimensional vector spaces $E$, $F$ and $G$ over $\mathbb{Z}/2$:

$E\xrightarrow{A}F\xrightarrow{B}G$

then we take transpose $A^t$ and $B^t$ and consider chain

$E\xleftarrow{A^t}F\xleftarrow{B^t}G$.

Is there any way to prove that $\ker(B)/Im(A)\cong \ker(A^t)/Im(B^t)$?

Note: This question is regarding isomorphism of homologies in proof of Poincare duality for Morse Homology, however I need purely algebraic result here. In the proof maps $A$ and $B$ are given by chain maps for Morse function $f$ while $A^t$ and $B^t$ correspond to chain maps between morse complexes of function $-f$.

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    $\begingroup$ Have you tried arguing these are vector spaces over the same field, of the same dimension and therefor isomorphic? $\endgroup$ – jMdA Jan 23 at 3:16
  • $\begingroup$ Yes, but I can't see why dimension would be the same. There would have to be some relation between $\ker(A^t)$ and $Im(A)$ which implies that? $\endgroup$ – OSBM Jan 23 at 3:20
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    $\begingroup$ @OSBM Row rank is the column rank and vice versa. $\endgroup$ – mathematics2x2life Jan 23 at 3:27
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    $\begingroup$ Using this result I have that dimension of $Im(A^t)$ is equal to co-dimension of $\ker(A)$, and other way around by rank nullity. Hence, working in $\mathbb{Z}/2$, I have that $\dim \ker(A^t)/Im(B^t) = n- \dim Im(A) - (n-\dim ker(B) )= \dim ker(B) -\dim Im(A) = \dim \ker(B)/Im(A)$ Then result follows by @jMdA argument. $\endgroup$ – OSBM Jan 23 at 3:39
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    $\begingroup$ Doesn't. Poincare want a natural isomorphism? So just comparing dimensions might not suffice. I would dave tried to uno left exactness of $hom(k,_)$ $\endgroup$ – Enkidu Jan 23 at 7:53

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