Hatcher Exercise 2.2.38 
I'm struggling to show exactness at $C_n\oplus D_n$. Let's take $(x,y)\in C_n\oplus D_n$ in the kernel of $C_n\oplus D_n\to E_n$, i.e. the pushforwards $x', y'$ into $E_n$ resp. satisfy $x' + y' = 0$. We need to find a pre-image of $(x,y)$ in $B_n$.
By exactness of the bottom, $y'$ pushed forward to $A_{n-1}$ gives $0$ (two steps in an exact sequence). Then $x'=-y'$ also goes to $0$, hence $x$ goes to $0$ by commutativity. Then $x$ has a pre-image $x_1\in B_n$.
I'm stuck here. Am I on the right track? How do I proceed?
 A: You're indeed on the right track. $x_1$ already has the correct image in $C_n$, but unfortunately maybe not in $D_n$, so we need to correct $x_1$ with something, such that the image in $C_n$ stays untouched and the image in $D_n$ is the right one.
Call the image of $x_1$ in $D_n$ $d$. The image of $d$ in $E_n$ is $x'=-y'$, hence $d+y$ is zero in $E_n$, so it comes from $A_n$. Now try to proceed using the element in $A_n$ to correct $x_1$ in such a way it maps to $y$ in $D_n$.
Comment my answer, if you are not able to finish it by yourself and I'll help you. 
A: Homological Algebra is a game of arrows, if you can fully make use of the arrows, you can prove theorems.
First of all, let us define $\sigma_{AB}, \sigma_{AD}, \sigma_{BD}, \sigma_{BC}, \sigma_{CE}, \sigma_{DE}$ be maps of corresponding arrows. Let   $$B_n \xrightarrow{\phi} C_n \oplus D_n\xrightarrow{\epsilon}E_n$$
with  $$\phi=\sigma_{BC} \oplus (-\sigma_{BD}),\\ \epsilon=\sigma_{CE} \oplus \sigma_{DE}$$ 
Now, it is obvious that $ker\epsilon=\{(c,d)\in C_n \oplus D_n: \sigma_{CE}(c)+\sigma_{DE}(d)=0   \}$,and $Im\phi \subset ker\epsilon$, what we should do is to show that $Im\phi \supset ker\epsilon$.
Step1, let $(c,d)\in ker\epsilon$, we have $\sigma_{CE}(c)=-\sigma_{DE}(d)=\sigma_{DE}(-d)$, and then apply $\sigma_{EA}$, we have $0=\sigma_{EA}\circ\sigma_{DE}(-d)=\sigma_{EA}\circ\sigma_{CE}(c)=\sigma_{CA}(c)$ by the communtativity of the diagram. Now by the exactness of above complex, we have $\exists b\in B_n$, s.t. $\sigma_{BC}(b)=c.$
Step2, consider the element $\Delta=\sigma_{BD}(b)+d\in D_n$, apply $\sigma_{DE}$ to $\Delta$, we have $\sigma_{DE}(\Delta)=\sigma_{DE}(\sigma_{BD}(b)+d)=\sigma_{DE}\circ\sigma_{BD}(b)+\sigma_{DE}(d)=\sigma_{CE}\circ\sigma_{BE}(b)+\sigma_{DE}(d)=\sigma_{CE}(c)+\sigma_{DE}(d)=0$. Therefore $\Delta\in ker\sigma_{DE}=Im\sigma_{AD}$, and so $\exists a\in A_n$, s.t. $\sigma_{AD}(a)=\Delta$
Step3, consider the element $\beta=\sigma_{AB}(a)-b\in B_n$. You can check that $\beta$ is the element such that $\phi(\beta)=(c,d)$.
Q.E.D.
