The Short Five Lemma

Let $R$ be a ring (with unity) and suppose we have a morphism of short exact sequence of $R$ -modules:

Prove:

if $h_{1}$ and $h_{3}$ are injective, then $h_{2}$ is injective
if $h_{1}$ and $h_{3}$ are surjective, then $h_{2}$ is surjective

Hints

For the injectivity of $h_{2}$ , start with an element $y \in Y$ with $h_{2} (y) = 0$ . Chase that element around the diagram to eventually find an element $x \in X$ with $f (x) = y$ . Then chase $x$ around the diagram to show $x = 0$ and hence $y = 0$ . You will use:

Commutativity of the right square
Injectivity of $h_{3}$
Exactness of the top row at $Y$
Commutativity of the left square
Exactness of the bottom row at $X^{'}$
Injectivity of $h_{1}$

For the surjectivity of $h_{2}$ , start with an element $y^{'} \in Y$ . The goal is to show there is some $y \in Y$ with $h_{2} (y) = y^{'}$ . First show there is some $y_{1} \in Y$ such that $g^{'} (h_{2} (y_{1})) = g^{'} (y^{'})$ . Then prove there is an element $x \in X$ such that $f^{'} (h_{1} (x)) = y^{'} - h_{2} (y_{1})$ . Finally, show that the element $f (x) + y_{1}$ maps to $y^{'}$ under $h_{2}$ . You will use:

Surjectivity of $h_{3}$
Exactness of the top row at $Z$
Commutativity of the right square
Exactness of the bottom row at $Y^{'}$
Surjectivity of $h_{1}$
Commutativity of the left square