Question concerning exact sequences of $R$-modules Let $R$ be a commutative ring. I would like to prove the following two assertions :

(1) If $0 \longrightarrow X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y$ is an exact sequence of $R$-modules, then $ker(\beta) \simeq X$.
(2) If $X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \longrightarrow 0$ is an exact sequence of $R$-modules, then $coker(\alpha) \simeq Y.$

At the moment I only see that by exactness we have $im(\alpha) = Z = ker(\beta)$.
Can someone explain me how to proceed ?
Thanks for your help.
 A: Hint:


*

*What is $\ker(\alpha)$? What is the relation between $\text{im}(\alpha)$ and $X$?

*$\text{coker}(\alpha)=Z/\text{im}(\beta)$ by definiton. If you can show that $\beta$ is surjective, you can use the 1st isomorphism theorem.

A: For (1):
The exact sequence
$0 \longrightarrow X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \tag 1$
implies that
$\ker \alpha = \text{Im}(0) = 0, \tag 2$
thus $\alpha:X \to Z$ is injective, meaning 
$\alpha:X \to \text{Im} \; \alpha \subset Z \tag 3$
is an isomorphism of $X$ to $\alpha(X) = \text{Im} \; \alpha $; now exactness of (1) yields
$\text{Im} \; \alpha = \ker \beta; \tag 4$
thus $\alpha$ is an isomorphism
$\alpha:X \simeq \ker \beta. \tag 5$
As for (2), the exact sequence
$X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \longrightarrow 0 \tag 6$
yields
$\text{Im} \; \beta = \ker (Y \to 0) = Y, \tag 7$
and
$\ker \beta = \text{Im} \; \alpha; \tag 8$
(7) means $\beta$ is surjective; thus
$Z / \ker \beta \simeq Y, \tag 9$
so by (8),
$Z / \text{Im} \; \alpha \simeq Y, \tag{10}$
which is precisely the statement
$\text{coker} \; \alpha \simeq Y; \tag{11}$
see here.
