The Short Five Lemma

For the injectivity of ${\tau }_2$, start with an element $y\in Y$ with ${\tau }_2(y)=0$. 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 ${\tau }_3$
• Exactness of the top row at $Y$
• Commutativity of the left square
• Exactness of the bottom row at $X^{\prime }$
• Injectivity of ${\tau }_1$

For the surjectivity of ${\tau }_2$, start with an element $y^{\prime }\in Y$. The goal is to show there is some $y\in Y$ with ${\tau }_2(y)=y^{\prime }$. First show there is some $y_1\in Y$ such that $g^{\prime }({\tau }_2(y_1))=g^{\prime }(y^{\prime })$. Then prove there is an element $x\in X$ such that $f^{\prime }({\tau }_1(x))=y^{\prime }-{\tau }_2(y_1)$. Finally, show that the element $f(x)+y_1$ maps to $y^{\prime }$ under ${\tau }_2$. You will use:

• Surjectivity of ${\tau }_3$
• Exactness of the top row at $Z$
• Commutativity of the right square
• Exactness of the bottom row at $Y^{\prime }$
• Surjectivity of ${\tau }_1$
• Commutativity of the left square