Short exact sequences envolving syzygies The following Lemma came from the paper Generalized Igusa-Todorov function and finitistic dimensions - D. Xu, in which is described by the author as a "basic homological fact" and is presented without proof. Maybe I'm blind, but I can't figure out how this can be proved easily. If someone can help and present the proof of the following Lemma I'll be really glad.
Lemma 4.1.
Let $A$ be an Artin algebra and $0\rightarrow X \rightarrow Y\rightarrow Z \rightarrow 0$ an exact sequence in $A$-mod. Then there are exact sequences 
$(1)$ $0\rightarrow \Omega_A(Z) \rightarrow X\oplus Q \rightarrow Y \rightarrow 0$ 
$(2)$ $0\rightarrow \Omega_A(Y) \rightarrow \Omega_A(Z)\oplus P \rightarrow Y \rightarrow 0$ 
$(3)$ $0\rightarrow \Omega_A(X) \rightarrow \Omega_A(Y)\oplus R \rightarrow \Omega_A(Z) \rightarrow 0$ 
where $P$, $Q$ and $R$ are projective $A$-modules. 
Notation: $\Omega_A(X)=\ker(f)$ where $f: P\rightarrow X$ is a projective cover. 
 A: This argument may have holes, and I would have put it in a comment, but I don't think it would fit.
Anyway, the same argument should work for (2) and (3) as for (1), so let's focus on (1). Consider the following diagram, where $Q \to Z$ is a projective cover:
$$
\begin{array}{ccccccccc}
& & 0 & & 0 & & 0 \\
& & \downarrow & & \downarrow & & \downarrow \\
0 & \to & \Omega Z & \to & X \oplus Q & \to & Y & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow \\
0 & \to & Q & \to & Y \oplus Q & \to & Y & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow \\
0 & \to & Z & \to & Z & \to & 0 & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow \\
& & 0 & & 0 & & 0
\end{array}
$$
The columns are exact, the bottom two rows are exact, and (here's where the hole may be) I think you can arrange for the diagram to commute. From that, you can conclude that the top row is exact.

I think the argument goes like this: start with the bottom row and the vertical maps from $Q$ and $Y$:
$$
\begin{array}{ccccccccc}
& & Q & & Y \\
& & \downarrow^p & & \downarrow^g \\
0 & \to & Z & = & Z & \to & 0 & \to & 0
\end{array}
$$
Since the right vertical map is surjective (because of the original short exact sequence) and $Q$ is projective, there is a map $h: Q \to Y$ making the square commute: that is, $gh=p$. Now I claim that this is an exact sequence:
$$
0 \to Q \xrightarrow{(h,1)} Y \oplus Q \xrightarrow{1-h} Y \to 0
$$
where the first map sends $q$ to $(h(q), q)$ and the second map sends $(y,q)$ to $y-h(q)$. Then this diagram commutes:
$$
\begin{array}{ccccccccc}
0 & \to & Q & \xrightarrow{(h,1)} & Y \oplus Q & \to & Y & \to & 0 \\
& & \downarrow^p & & \downarrow^{g+0} & & \downarrow \\
0 & \to & Z & = & Z & \to & 0 & \to & 0
\end{array}
$$
where the second vertical map sends $(y,q)$ to $g(y)$. This commutes because if $q \in Q$, then going down and then right gives $p(q)$, while going right and then down gives $q \mapsto (h(q), q) \mapsto g h(q) = p(q)$ by construction of $h$.
The kernel of the vertical map $(g,0)$ is $X \oplus Q$, and the kernel of $p$ is $\Omega Z$.
