Criteria for right exact sequence How do I prove:
$A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0$ is a exact sequence of modules if and only if for every $D$,  and $f:B\to D$ such that $f\alpha=0$, there exist a unique $f':C\to D$ such that $f=f'\beta$.
$\Longrightarrow$ is easy, I need help with the other direction.  There are a few choices for  $D$, such as $\ker\alpha$, $\ker\beta$, $image\ \alpha$ and $image\ \beta$
 but I can't find reasonable $f$ and $f'$ for any of them.
Attempt:
Exactness at $C$: let $D=C/\beta(B)$. $f=0:B\to D$, then both $q:C\to C/\beta(B)$ the quotient map, and $0:C\to C/\beta(B)$ makes the diagram commute.  By the uniqueness of $f'$ we must have $q=0$ which only happens if $\beta(B)=C$.
Exactness at $B$, $im(\alpha)\subset\ker\beta$, $im(\alpha)\supset\ker\beta$:  (will fill in later tomorrow)
 A: I do not see how the problem is correct as stated.  What I do agree with:


*

*As stated, the implication $\Rightarrow$ is true.  I agree with you here.  (I think the converse is false, though.)

*As stated, one can prove that $\beta$ is onto, just as you have.

*As stated, one can prove that $\ker \beta \subseteq  \mathrm{im}\alpha $.  Proof is below.


I cannot get $\mathrm{im}\alpha \subseteq  \ker \beta $ to complete exactness, and I think I have a counterexample.
Counterexample to problem as stated:  The following sequence is NOT exact yet it does fulfill your universal mapping property:  
$$
\mathbb{Z}_8 \stackrel{\alpha}{\to} \mathbb{Z}_4 \stackrel{1}{\to} \mathbb{Z}_4 \to 0
$$
where $\alpha$ is multiplication by $2$.  Note that we have $\beta = 1$ is onto and that $\ker \beta = \{0\}$ is an obvious subgroup of $\mathrm{im}\alpha =\{0,2\}$.  So far, everything works.  They are not equal, of course, so this isn't exact.
Let $f: \mathbb{Z}_4 \to D$ be any map with $f\alpha =0$.  This requirement forces $f(2)=0$ since $f(2)=f(\alpha(1)) =0$.  Likewise $f(k)=0$ when $k$ is $0, 2, 4$.  Obviously $f$ is dictated by what it does to $1$ and this must map to an element of $D$ of order 1, 2, or 4.  I'm not sure this much matters, though, since the ONLY map $f': \mathbb{Z}_4 \to D$ with $f' \circ \beta = f$ is $f'=f$ itself since $\beta=1$.  So unique fill-ins exist no matter what $f$ and $D$ are, so your statement holds.  Yet, the sequence isn't exact. 
Proof that  $\ker \beta \subseteq  \mathrm{im}\alpha $ with the original problem hypotheses:  Apply the hypotheses to $D = B/\mathrm{im}\alpha$
and $f=q:B \to B/\mathrm{im}\alpha$ the obvious quotient map.  Let $q': C \to 
B/\mathrm{im}\alpha$ be the unique fill-in whose existence is asserted by the assumptions.  You can check that, because of uniquess, $q'(c)$ must be defined as
$$
q'(c) = b + \mathrm{im}\alpha \text{ where } \beta(b)=c.
$$
The point is that this formula gives a well-defined homomorphism making the right triangle commute, so by uniqueness this must be the formula for $q'$ (this part needs $\beta$ to be onto, which you have already proven).  
Take any $x \in \ker \beta$.  Then $q'(\beta(x))$ must be $0  + \mathrm{im}\alpha$, but by definition this is $x  + \mathrm{im}\alpha$.  But $0  + \mathrm{im}\alpha = x  + \mathrm{im}\alpha$ implies $x \in \mathrm{im}\alpha$, so $\ker \beta \subseteq  \mathrm{im}\alpha $ as claimed.
If someone sees a mistake in ANY of this, please point it out.  This problem has really been bothering me. 
