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Let $0 \rightarrow M' \xrightarrow{f} M \xrightarrow{g} M'' \rightarrow 0$ by an exact sequence of modules. the following conditions are equivalent

  1. There exists a homomorphism $\phi: M'' \rightarrow M$ such that $g \circ \phi = $ id.

  2. There exists a homomorphism $\psi: M \rightarrow M'$ such that $\psi \circ f = $ id.

$(1) \implies (2)$. I try to show $f$ is injective and surjetive. Since the sequence $0 \rightarrow M' \xrightarrow{f} M$ is exact, $f$ is injective. Since $g$ has an inverse, $\ker(g) = 0$. Thus, since the sequence $M' \xrightarrow{f} M \xrightarrow{g} M''$ is exact, $f$ maps identically to $0$???

This last step I'm having some trouble getting around. If $f$ maps to $0$ it can't be surjective unless $M$ only contains $0$.

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    $\begingroup$ Why are you trying to show that $f$ is surjective? In fact, that's not even true for a general split exact sequence. $\endgroup$ Commented Jun 17, 2021 at 15:16
  • $\begingroup$ to show that $f$ has an inverse, it must be both surjective and injective is that not correct? $\endgroup$ Commented Jun 17, 2021 at 15:18
  • $\begingroup$ Also, $\ker g = 0$ is also not true for a general split exact sequence. Note that the compositions you've been given only show that $f$ has a left inverse and $g$ has a right inverse. Neither implies that $f$ or $g$ is bijective. To be very explicit: You are not given that $f$ (or $g$) has an inverse. $\endgroup$ Commented Jun 17, 2021 at 15:18
  • $\begingroup$ The prototypical example of a split exact sequence is a sequence of the form $$0 \to M \to M \oplus M' \to M' \to 0$$ where the arrows are the natural ones (inclusion and projection). In fact, an equivalent way of stating a sequence to be split exact is that it is isomorphic to the above in a nice way. Unless $M$ or $M' = 0$, the arrows above are not bijective. $\endgroup$ Commented Jun 17, 2021 at 15:21
  • $\begingroup$ Hint: take as $\psi$ the cokernel of $\varphi$. $\endgroup$ Commented Jun 17, 2021 at 16:02

1 Answer 1

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We will show that your statements 1. and 2. are equivalent to the following:

  1. There exists an isomorphism $\mu : M \to M' \oplus M''$ such that the following diagram with exact rows commutes

\begin{CD} 0 @>>> M' @>f>> M @>g>> M'' @>>> 0\\ @. @V\text{Id}_{M'}VV @V\mu VV @VV\text{Id}_{M''}V\\ 0 @>>> M' @>i>> M' \oplus M'' @>p>> M'' @>>> 0 \end{CD} Here, $i$ and $p$ are the obvious injection and projection respectively.

(3. $\implies$ 2.) Define $\psi := \text{pr}_{M'} \circ \mu$. Here, $\text{pr}_{M'} : M' \oplus M'' \to M'$ is defined by $(m',m'') \mapsto m'$. Then for $m' \in M'$ we have $$ (\psi \circ f)(m') = \text{pr}_{M'}(\mu(f(m'))) = \text{pr}_{M'}(i(m')) = \text{pr}_{M'}(m',0) = m', $$ where we used that the left square in the diagram commutes.

(3. $\implies$ 1.) Define $\phi(m'') := \mu^{-1}(0,m'')$ for $m'' \in M''$. Then we have $$ (g \circ \phi)(m'') = g(\mu^{-1}(0,m'')) = p(0,m'')=m'', $$ where again we used that the right square commutes, and that $\mu$ is bijective.

(2. $\implies$ 3.) Let $\mu(m) := (\psi(m),g(m))$. You can check that the diagram does indeed commute. By the five lemma, $\mu$ is an isomorphism, since $\text{Id}$ is always bijective.

(1. $\implies$ 3.) Define $\nu : M' \oplus M'' \to M$ by $\nu(m',m'') = f(m')+\phi(m'')$. You can check that the following diagram then commutes: \begin{CD} 0 @>>> M' @>f>> M @>g>> M'' @>>> 0\\ @. @A\text{Id}_{M'}AA @A\nu AA @AA\text{Id}_{M''}A\\ 0 @>>> M' @>i>> M' \oplus M'' @>p>> M'' @>>> 0 \end{CD} Again, by the five lemma we have that $\nu$ is an isomorphism, so 3. follows.

Now we have shown that 1. and 2. are equivalent to 3., so in particular 1. and 2. are equivalent. Hope this helps!

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