In the proof of lemma 3.2 in this PDF, it said:
Let $0 → A → A′ → A′′ → 0$ be a short exact sequence. $P′′_{*}→ A′′$ be a projective resolution. (i.e. $ ··· P_{1}′′→ P_{0}′′ → A′′ → 0$ is exact with $P_{i}$ projective)
Now let $α′\in A′$. If $α′ \notin Ker(A′ → A′′)$ then it has nonzero image $α′′ ∈ A′′$, and there is some $p′′ ∈ P_{0}′′$ mapping onto $α′′$. By commutativity, this $p′′$ maps onto $α′$.
Why $p′′$ must maps onto $α′$ but not other elements which also has $α′′$ as its image?
I think something is off here. Certainly the claim that every element of $A'$ is in the image of $P_0$ or the image of $P_0''$ seems suspicious -- there are many more elements in the direct sum, and I think if you took the original exact sequence to be, say, a split exact sequence of free modules, then it seems as if the process described would just spit a copy of that original column, and then the claim is false.
But it doesn't take a lot of patching up. If $x$ is the image of $p''$ in $A'$ then $x - \alpha'$ maps to $0$ in $A''$, so you can add an element of $P_0$ to $p''$ to get something in $P_0 \oplus P_0''$ that maps to $\alpha'$.
"By commutativity of (3)."
The point is that, since $P''$ is projective and $A' \to A''$ is surjective, we've defined a map $P'' \to A'$ making the triangle commute. That is, the composition $P'' \to A' \to A''$ is the same as the map $P'' \to A''$, and we have defined the map $P'' \to A'$ so that it works.
