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Maybe I'm underestimating your problem, but it seems Mikael above is right.

In your example you define $q:f_1\to f_2$, so if it's a kernel of some other map r, then $r$ must have $f_2$ for domain. $q$ can't possibly be a kernel of $p$, as the composition $pq$ does not make sense.

Categorically http://en.wikipedia.org/wiki/Kernel_(category_theory)

Given a map $f: X \to Y$, a kernel is another map $k:K \to X$ satisfying blah blah.

So

Now, if X and Y are complexes you have a criterion to check wether $k$ is a kernel: checking the components $k^n$ (but you already must have a chain map $k$ to begin with).

1

Maybe I'm underestimating your problem, but it seems Mikael above is right.

In your example you define $q:f_1\to f_2$, so if it's a kernel of some other map r, then $r$ must have $f_2$ for domain. $q$ can't possibly be a kernel of $p$, as the composition $pq$ does not make sense.

Categorically http://en.wikipedia.org/wiki/Kernel_(category_theory)

Given a map $f: X \to Y$, a kernel is another map $k:K \to X$ satisfying blah blah.

So if X and Y are complexes you have a criterion to check wether $k$ is a kernel: checking the components $k^n$ (but you already must have a chain map $k$ to begin with).
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