# Isomorphism of $\operatorname{top}f$ implies $f$ is a projective cover.

I'm stuck on the following and could use a hint.

Let $f:P\longrightarrow M$ be a map of finite dimensional modules over a finite dimensional algebra $A$ (over probably an algebraically closed field $K$), with $P$ projective. $f$ induces a map $\operatorname{top}f:\operatorname{top}P\longrightarrow \operatorname{top}M$ where $\operatorname{top}M = M/\operatorname{rad}M$ and $\operatorname{rad}M$ is the Jacobson radical.

Show that $\operatorname{top}f$ an isomorphism implies $f:P\longrightarrow M$ is a projective cover.

Don't really know where to start.

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Let $\pi_P:P\to P/\mbox{rad}P$ and $\pi_M:M\to M/\mbox{rad}M$ be the natural projections. The induced homomorphism $\mbox{top}f$ is such that $\pi_M\circ f=\mbox{top}f\circ\pi_P$. If $y\in M$ then, being $\mbox{top}f$ surjective, $y+\mbox{rad}M=\mbox{top}f(x+\mbox{rad}P)=f(x)+\mbox{rad}M$, for some $x\in P$. This proves that $M=\mbox{rad}M+\mbox{Im}f$. Since $M$ is finite dimensional, $\mbox{rad}M$ is superfluous in $M$, therefore $M=\mbox{Im}f$, i.e. $f$ is an epimorphism.
On the other hand, if $f(x)=0$ then $\mbox{top}f(x+\mbox{rad}P)=f(x)+\mbox{rad}M=0$, so being $\mbox{top}f$ a monomorphism we have $x\in\mbox{rad}P$. Therefore $\mbox{Ker}f\leq\mbox{rad}P$. Again, since $P$ is finite dimensional, $\mbox{rad}f$ is superfluous, so $\mbox{Ker}f$ is also superfluous.
We conclude that $f:P\to M$ is a projective cover.