# Isomorphism $\ker(B)/Im(A)\cong \ker(A^t)/Im(B^t)$ for chain of linear maps.

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 $$-f$$.

• Have you tried arguing these are vector spaces over the same field, of the same dimension and therefor isomorphic? – jMdA Jan 23 at 3:16
• 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? – OSBM Jan 23 at 3:20
• @OSBM Row rank is the column rank and vice versa. – mathematics2x2life Jan 23 at 3:27
• 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. – OSBM Jan 23 at 3:39
• Doesn't. Poincare want a natural isomorphism? So just comparing dimensions might not suffice. I would dave tried to uno left exactness of $hom(k,_)$ – Enkidu Jan 23 at 7:53