For an exact sequence $0\to M_1\overset{f_1}\to\cdots\overset{f_r}\to M_r\to0$ is it true that $l(M_i)-l(M_{i+1})=l(\ker(f_i))-l(\ker(f_{i+1}))$?

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For an exact sequence $0\to M_1\overset{f_1}\rightarrow\cdots\overset{f_r}\rightarrow M_r\to0$ is it true that $l(M_i)-l(M_{i+1})=l(\ker(f_i))-l(\ker(f_{i+1}))$?

$M_i$s are modules and $l(M_i)$ denotes the length of a module. Each module is of finite composition length. I couldn't fit it into the title, but the exact sequence also includes $0\rightarrow M_1$ at the beginning and $M_r\rightarrow 0$ at the end.

From the first isomophism theorem, we know that

$$M_i/\ker(f_i)\cong im(f_i)$$ and, from the definition of cokernel, we know that $$coker(f_i)=M_{i+1}/im(f_i)$$

Taking lengths, we get $$l(M_i)-l(\ker(f_i))=l(M_{i+1})-l(coker(f_i)).$$ This is almost in the form I want it to be, except I have $coker(f_i)$ instead of $\ker(f_{i+1})$. Why should they be equal?

I can see that $$coker(f_i)=M_{i+1}/im(f_i)=M_{i+1}/\ker(f_{i+1})\cong im(f_{i+1}),$$ but that's not really helping me.

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Because of:
$l(coker(f_i))=l(M_{i+1})-l(\ker(f_{i+1}))$ and $l(M_i)-l(\ker(f_i))=l(M_{i+1})-l(coker(f_i))$

We have: $$l(M_i)=l(\ker(f_i))+l(\ker(f_{i+1}))$$
And $$l(M_{i+1})=l(\ker(f_{i+1}))+l(\ker(f_{i+2}))$$
Hence: $$l(M_i)-l(M_{i+1})=l(\ker(f_i))-l(\ker(f_{i+2}))$$

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Aside from some problems with the question (lengths not necessarily being finite, making it problematic to take differences in general), the answer is still no. For instance, consider $$ M \rightarrow M \rightarrow 0 $$ where $M$ is some nonzero module of finite length, and with the map $M \rightarrow M$ the identity. Then $\ker(M\to 0)=M$, $\ker(M\to M)=0$, and $$l(M)-l(M)=0 \ne l(\ker(f_i)) - l(\ker(f_{i+1})) = -l(M).$$

In general, since $\operatorname{im}(f_i)=\ker(f_{i+1})$, you are trying to relate $l(M_i)$, $\ker(f_i)$, $\operatorname{im}{f_i}$, and $l(M_{i+1})$. But you can't expect a relation between those. For instance, you can always replace $M_{i+1}$ by a bigger module, as long as you change the sequence to the right so that it remains exact.