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how to prove that:

$\mathrm{Hom}(\varinjlim X_i,Y) \cong \varprojlim \mathrm{Hom}( X_i,Y)$?

Added(05/10/12)

It should be $Y$, instead of $Y_i$, and thanks for the answers.

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    By the universal property of the limit, homomorphisms from the inductive limit correspond to projective families of morphisms from the constituents of the limit. (Just like a homomorphism from a coproduct is equivalent to a family/product of homomorphisms from the cofactors).2012-05-10
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    it should be $Y_i$2012-05-10
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    @ArturoMagidin Sorry I don't know what you mean2012-05-10
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    Shouldn't it read $\mathrm{Hom}(\varinjlim X_i, Y)$?2012-05-10
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    Your indices don't make sense. The first index $i$ on the left is bound by the $\varinjlim$, but the $i$ in $Y_i$ is free; but both indices on the right hand side are bound by the $\varprojlim$2012-05-10
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    Assuming it should be $Y$, the isomorphism you are asking is just a restatement of the universal property of $\varinjlim$: a map from $\varinjlim X_i$ corresponds uniquely to a family of maps from the $X_i$ that respect the inductive structure. Write out what "respect the inductive structure" means and you'll discover that you have a *projective* family of morphisms. This is completely analogous to the fact that $\mathrm{Hom}(\coprod X_i, Y) \cong \prod\mathrm{Hom}(X_i,Y)$, because given $\{f_i\}$ on the right, it induces a unique $f$ on the left. Same thing here.2012-05-10

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