Diagram Chase: Ascending Chain of Kernels I have stumbled across what should be a routine diagram chase which I'm struggling with.
The question is as follows:
Let $f \colon A \to B$ be a map of Abelian groups. Let $p_A \colon A \to A$ and $p_B \colon B \to B$. Assume:

*

*$ f \circ p_A = p_B \circ f $.

*$A = \bigcup_{i \geq 1} \ker(p_A^i)$ and $B = \bigcup_{i \geq 1} \ker(p_B^i)$.

*$f \colon \ker(p_A) \to \ker(p_B)$ is an isomorphism.

*$f \colon \text{coker}(p_A) \to \text{coker}(p_B)$ is a monomorphism.

We want to show that $f$ is an isomorphism.
The book draws some diagrams (see below!) and claims the induction is straightforward, but I cannot even seem to show that $\ker(p_A^2) \to \ker(p_B^2)$ is an isomorphism.

EDIT(s): The book suppresses the subscript on the $p$. One should interpret the upper row as dealing with $p_A$ and the lower row as dealing with $p_B$. Presumably one should successively show that $f|_{\ker(p_A^i)} \colon {\ker(p_A^i)} \to {\ker(p_B^i)}$ is an isomorphism, although this is not stated explicitly.
There seems to be a typo: $\ker(p)/p\ker(A)$ doesn't make sense, and neither does $\ker(p)/pB$ now that I think about it! Needless to say, I'm even more confused!
 A: I think this is what was intended:
$\newcommand{cok}{\text{cok}}$
$\require{AMScd}$
The proof is by induction; we claim that for each $n \geq 1$, the map
$\ker (p^{n}) \to \ker (p^{n})$ is an isomorphism and the map $\cok
(p^{n}) \to \cok (p^{n})$ is a monomorphism (with the apparent subscripts omitted — I believe that in the original setting, they just referred to multiplying by a prime number $p$, for what that's worth). We know that these are true when $n=1$.
The inductive step: for each $n \geq 1$, the following are commutative
diagrams with exact rows; the vertical maps are all given by $f$:
\begin{CD}
0 @>>> \ker(p^{n-1}) @>>> \ker(p^{n}) @>{p^{n-1}}>> \ker(p) @>>> \ker(p)/p^{n-1} \ker(p^{n})
@>{\text{1-1}}>> \cok(p^{n-1}) \\
@. @VV{\cong}V @VVV @VV{\cong}V @VVV @VV{\text{1-1}}V\\
0 @>>> \ker(p^{n-1}) @>>> \ker(p^{n}) @>{p^{n-1}}>> \ker(p) @>>> \ker(p)/p^{n-1} \ker(p^{n})
@>{\text{1-1}}>> \cok(p^{n-1})
\end{CD}
and
\begin{CD}
\ker (p^{n-1})/p \ker (p^{n}) @>>> \cok (p) @>{p^{n-1}}>> \cok (p^{n}) @>>> \cok (p^{n-1}) @>>> 0 \\
@VVV @VV{\text{1-1}}V @VVV @VV{\text{1-1}}V \\
\ker (p^{n-1})/p \ker (p^{n}) @>>> \cok (p) @>{p^{n-1}}>> \cok (p^{n}) @>>> \cok (p^{n-1}) @>>> 0\\
\end{CD}
The first diagram and the five lemma imply that $\ker (p^{n}) \to \ker
(p^{n})$ is an isomorphism. Then the left-hand vertical map in the
second diagram is an isomorphism, so one of the four lemmas (as
described on Wikipedia's Five lemma page) implies that $\cok (p^{n})
\to \cok (p^{n})$ is a monomorphism.
