How to choose a left-add$(X)$-approximation with a certain property Let $A$ be an artin algebra and $X,Y$ in mod-$A$. Suppose $0\rightarrow Y \stackrel{\alpha}{\rightarrow} X^n\stackrel{\beta}{\rightarrow} X^m$ is exact.
Set $C:=Coker(\alpha)$ (as module) and $c:=coker(\alpha)$ (as a map).

Why is it then possible to choose a left-$add(X)$-approximation $\varphi:C\rightarrow X^{\widehat{m}}$ (this means that the induced map Hom$_A(X^{\widehat{m}},Z)\rightarrow$Hom$_A(C,Z)$ is surjective for all $Z\in add(X)$) with the property that the sequence $0\rightarrow Y \stackrel{\alpha}{\rightarrow} X^n\stackrel{\gamma}{\rightarrow} X^{\widehat{m}}$, with $\gamma:=\varphi\circ c$, is still exact?

Thank you for your effort.

First, note that there always exists a left-$add(X)$-approximation $\varphi:C\to X^{\widehat{m}}$ for any module $C$ in mod-$A$.  This comes from the fact that $Hom_A(C,X)$ is a finitely generated module over the (implicit) base ring $k$; if $f_1, \ldots, f_r$ are generators, then the map given in matrix form by $(f_1, \ldots, f_r)^t$ from $C$ to $X^r$ is an approximation.
Next, under the assumption of the question, we can prove that this $\varphi$ is injective.  Indeed, since the sequence $0\to Y\stackrel{\alpha}{\to} X^n \stackrel{\beta}{\to} X^m$ is exact, we have that $C$ is isomorphic to the image of $\beta$.  As a consequence, there is an injection $j:C\to X^m$.  Since $\varphi$ is a left-$add(X)$-approximation, $j$ factors through $\varphi$ (meaning that there is a $g:X^{\widehat{m}}\to X^m$ such that $g\varphi=j$).  Since $j$ is injective, $\varphi$ has to be as well.
Finally, we only have to concatenate the exact sequences $0\to Y \stackrel{\alpha}{\to} X^n \to C \to 0$ and $0\to C \stackrel{\varphi}{\to} X^{\widehat{m}}$ to obtain the exact sequence $0\to Y \stackrel{\alpha}{\to} X^n \to X^{\widehat{m}}$.
