For a pure-injective module $M$ does the property "$\operatorname{Hom}(-,M)$ is surjective" commute with certain limits?

Question

$\DeclareMathOperator\Hom{Hom}$Let $M$ be a pure-injective module. Then $\Hom(\varphi,M)$ is surjective for a pure-mono $\varphi$. It is well-known that $\varphi$ is a direct limit of split monos $\varphi_i$ between finitely presented modules. Clearly all $\Hom(\varphi_i, M) $ are surjective. This motivates the following question:

Let $M$ be a pure-injective module, $\psi_i$ a directed system of monos between finitely presented modules and $\psi$ the direct limit of this system. If all $\Hom(\psi_i, M)$ are surjective, does it follow that $\Hom(\psi, M)$ is surjective?

Answer 1

$\DeclareMathOperator\Hom{Hom}$The answer seems to be positive.

Let $\psi_i: X_i \rightarrow Y_i$. Consider the directed system of exact sequences

$$ 0 \longrightarrow K_i \longrightarrow X_i\otimes- \xrightarrow{\psi_i \otimes -} Y_i \otimes -$$

in the functor category. Its direct limit equals an exact sequence

$$ 0 \longrightarrow K \longrightarrow X\otimes- \xrightarrow{\psi \otimes -} Y \otimes -$$

since direct limits are exact and tensors commute with them. Now $M$ is pure-injective, so the functor $M\otimes-$ is injective and there exists an inverse system of short exact sequences

$$ \Hom(Y_i\otimes-,M\otimes-) \xrightarrow{\Hom(\psi _i \otimes - , M\otimes -)} \Hom(X_i\otimes-, M\otimes-)\longrightarrow \Hom(K_i, M \otimes-)\longrightarrow 0. $$

The natural isomorphism $\Hom(\psi _i \otimes - , M\otimes -)\cong \Hom(\psi_i, M)$ implies $\Hom(K_i, M \otimes-) = 0$. Because $\Hom(-,-)$ turns direct limits in the first argument into inverse limits, taking the inverse limit of the inverse system yields

$$ \Hom(Y\otimes-,M\otimes-) \xrightarrow{\Hom(\psi \otimes - , M\otimes -)} \Hom(X\otimes-, M\otimes-)\longrightarrow \Hom(K, M \otimes-) = 0.$$

Hence $\Hom(\psi, M)\cong \Hom(\psi \otimes - , M\otimes -) $ is surjective.