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how to prove that:

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


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). – Arturo Magidin May 10 '12 at 5:30
it should be $Y_i$ – Jr. May 10 '12 at 5:39
Shouldn't it read $\mathrm{Hom}(\varinjlim X_i, Y)$? – martini May 10 '12 at 5:48
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 $\varprojlim$ – Arturo Magidin May 10 '12 at 5:49
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. – Arturo Magidin May 10 '12 at 5:51

as said in my comment above, I suppose you meant to show $$\mathrm{Hom}(\varinjlim X_i, Y) \cong \varprojlim\mathrm{Hom}(X_i, Y),$$ as your indices as written don't make sense.

What Arturo meant is, that we should use the universal property of $\varprojlim$, that is we show that $\mathrm{Hom}(\varinjlim X_i, Y)$ has the properties of a limit of $\mathrm{Hom}(X_i, Y)$. For $i \in I$ let $\rho_i\colon X_i \to \varinjlim X_i$ the canonical morphism, we get a map $\rho_{i}^\star\colon \mathrm{Hom}(\varinjlim X_i, Y) \to \mathrm{Hom}(X_i, Y)$. For a morphism $f\colon i \to j$ we have an $X_f \colon X_i \to X_j$ an by the properties of the colimit we have $\rho_j \circ X_f = \rho_i$, which gives $\rho_i^\star = X_f^\star \circ \rho_j^\star$ by the functorial properties of $\star$. So the $\rho_i^\star$ interact well with the morphisms in $I$, we now have to show that there are universal in such. So let $\sigma_i\colon Z \to \mathrm{Hom}(X_i, Y)$ maps with $\sigma_i = X_f^\star \circ \sigma_j$ for each morphism $f\colon i \to j$ in $I$. For $z \in Z$ we have a family $\sigma_i(z) \colon X_i \to Y$ of morphisms. If $f\colon i \to j$ we have $\sigma_j(z) \circ X_f = X_f^\star\bigl(\sigma_j(z)\bigr) = (X_f^\star\circ \sigma_j)(z) = \sigma_i(z)$. By the universal property of $\varinjlim X_i$ there is a unique $\phi(z): \varinjlim X_i \to Y$ with $\phi(z) \circ \rho_i = \sigma_i(z)$ for all $i$. This gives a $\phi\colon Z \to \mathrm{Hom}(\varinjlim X_i, Y)$ with $\rho_i^\star \circ \phi = \sigma_i$ for all $i$. If $\psi\colon Z \to \mathrm{Hom}(\varinjlim X_i, Y)$ also has this property, we have for $z \in Z$ that $\psi(z) \circ \rho_i = \sigma_i(z)$ for all $i$ and by uniqueness of $\phi(z)$ we have $\psi(z) = \phi(z)$. As $z$ was arbitrary, $\phi = \psi$.

So $\mathrm{Hom}(\varinjlim X_i, Y)$ has the properties of a limit of $\mathrm{Hom}(X_i, Y)$ and therefore $$\mathrm{Hom}(\varinjlim X_i, Y) \cong \varprojlim\mathrm{Hom}(X_i, Y).$$

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