Let $E$ be a module over a ring. Let $\{ M_i \}$ be a directed family of modules. If $E$ is finitely generated, then the natural homomorphism $$\varinjlim \mathrm{Hom}(E, M_i) \to \mathrm{Hom}(E, \varinjlim M_i)$$ is injective.
I don't see the assumption that $E$ is finitely generated could really help here. Any hint would be appreciated. Thanks!
For $i<j$, denote the homomorphism in the direct family by $f_{ij}:M_{i}\rightarrow M_{j}$.
Consider an element $f$ in $\varinjlim \mathrm{Hom}(E,M_{i})$, it can be represented by some $f:E\rightarrow M_{i_{0}}$. If $f$ is contained in the kernel of the natural homomorphism:$\varinjlim \mathrm{Hom}(E,M_{i})\rightarrow \mathrm{Hom}(E,\varinjlim M_{i})$, we can see for each $x\in E$, the element $f(x)$ equals to zero in the direct limit $\varinjlim M_{i}$. This means there exists some $i=i(x)>i_{0}$ such that $f_{i_{0},i}(f(x))=0$.
Hence when $M$ is finitely generated, we can choose some generators $x_{1},\cdots,x_{n}$ of $M$ and then the index $i=\max\{i(x_{1}),\cdots,i(x_{n})\}$ satisfies $f_{i_{0},i}\circ f=0$ as a homomorphism from $E$ to $M_{i}$. This implies that $f=0$ in the direct limit of $\mathrm{Hom}(E,M_{i})$.
If $E$ is not finitely generated, we can construct the following counterexample: let $E$ be $k[\{T_{i}\}_{i\in I}],$ where $I$ is the index set of $\{M_{i}\}.$ Fix some $i_{0}$ and for each $i>i_{0}$, find one element $y_{i}\in M_{i_{0}}$ such that $f_{i_{0},j}(y_{i})\neq 0\in M_{j}$ for all $j<i$ and $y_{i}$ is killed by some $f_{i_{0},k},k>i$ (of course we can construct such a family of $\{M_{i}\}$). Define $f:E\rightarrow M_{i_{0}}$ that maps each $T_{i}$ to $y_{i}$ for $i>i_{0}$. This homomorphism is contained in the kernel of the natural map because each $y_{i}$ is zero in $\varinjlim M_{i}$. However by our construction this $f$ can not be killed by any $f_{i_{0},j}$.