Where is the mistake? (derived functors )

Question

Assume $pd(M) =n \leq \infty$ for a left $R$-module. I then have to show there exists a free module $F$ such that $Ext_{R}^{n}(M,F) \neq 0 $.

I have tried these steps and obtained a contradiction:

1. From $pd(M) =n \leq \infty$ I can conclude that there exists an $R$-module $F$ such that $Ext_{R}^{n}(M,F) \neq 0$ and $Ext_{R}^{n+1}(M,F) = 0$.

2. I write an injective resolution of $F$ as: \begin{equation} 0\rightarrow F \xrightarrow{\mu} E^{0} \xrightarrow{d_{0}} E^{1}\xrightarrow{d_{1}} \dots\end{equation}

3. Then I conclude from Rotman corollary 6.42: $0 = Ext_{R}^{n+1}(M,F) \cong Ext_{R}^{n}(M,\text{im }\mu) $

4. Since $Ext_{R}^{n}(M,F) \neq 0$ and $Ext_{R}^{n}(M,\text{im }\mu) \cong 0$ we have $\text{im }\mu \ncong F$

5. Using the first isomorphism theorem the kernel of $\mu$ cannot be 0

6. Hence the injective resolution is not exact. This is a contradiction.

   Where is the mistake?