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I read that $\lim\limits_{\longleftarrow}\mathrm{Hom}(N_j,M)\cong\mathrm{Hom}(\lim\limits_{\longrightarrow}N_j,M)$. I was wondering if we can write $\lim\limits_{\longrightarrow}\mathrm{Hom}(N_j,M)$ as $\mathrm{Hom}(X,M)$ for some $X$. Do you know if we can? I would be happy also if you can give me only a reference.

(Here I'm talking of modules over commutative rings, we can also suppose that the $N_j$'s and $M$ are finitely generated, we can also add some other conditions if you wish, maybe noetherianity).

I also read that $\mathrm{Hom}(M,\lim\limits_{\longleftarrow}N_j)\cong\lim\limits_{\longleftarrow}\mathrm{Hom}(M,N_j)$. I was wondering if under some hypothesis $\mathrm{Hom}(M,\lim\limits_{\longrightarrow}N_j)\cong \lim\limits_{\longrightarrow}\mathrm{Hom}(M,N_j)$. I would be happy if you can help me even in only one of those questions.

EDIT: what about if $M$ is not even finitely generated?

EDIT: Suppose that the transition maps in $\{N_j\}$ are injective. We can see $M$ as a direct limit of finitely generated modules. Is it true that $\lim\limits_{\longleftarrow_n}\lim\limits_{\longrightarrow_j}Hom(M_n,N_j)$=$\lim\limits_{\longrightarrow_j}\lim\limits_{\longleftarrow_n}Hom(M_n,N_j)$? If this is true then we can drop the hypothesis $M$ finitely generated in the answer of Matt E.

EDIT: what about $\mathrm{Hom}(\lim\limits_{\longleftarrow}\;M_n,N)$?

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    Just to make a link about the first point you have read about: http://math.stackexchange.com/questions/143356/relation-between-inductive-and-projective-limits/143371#1433712015-06-29

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There is a natural isomorphism $\varinjlim Hom(M,N_j) \cong Hom(M,\varinjlim N_j)$ for all filtered direct systems $N_j$ if and only if $M$ is finitely presented. If you assume that the transition maps in the system $\{N_j\}$ are injective, then finitely generated is enough. (See this answer.)

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    if you look to the answer of Pierre-Yves Gaillard to the question you linked it seems that inverse limits and direct limits as written in my question commute (or maybe I'm misunderstanding his answer)2012-11-01