# In one category $C$ Show that $\lim\coprod F_k \cong \prod \lim F_k$

I have some doubts with this problem because The concept of Limit is confusing for me:

"If one category $J$ is a disjoint union (coproduct) $\coprod_{k}F_{k}$ of categories $J_{k}$ for index $k$ in some set $K$, with $I_{k}:J_{k}\rightarrow J$ the injection of the coproduct, then each functor $F:J\rightarrow C$ determines Functors $F_{k}=FI_{k}:J_{k}\rightarrow C$.

Prove that $\lim F\cong \prod_{k} \lim F_{k}$ if the limit of the right exist."

$(1)$ For $\lim F$ we have morphism $v_{i}: \lim F\rightarrow F(i)$ such that for $u:i\rightarrow j$ we have that $v_{j}=F(u)\circ v_{i}$.

$(2)$ On other hand for each $\lim F_{k}$ we have morphism $v^{i_{k}}_{k}: \lim F_{k}\rightarrow F_{k}(i_{k})$ such that for $u_{k}:i_{k}\rightarrow j_{k}$ we have that $v^{j_{k}}_{k}=F_{k}(u_{k})\circ v^{i_{k}}_{k}$ (I am sorry for the notation i couldn't think on another notation )

My idea is if i fix $k$ and take $i_{k}\in J_{k}$ for $(1)$ we have a collection of morphism $v_{i_{k}}:\lim F\rightarrow F(i_{k})=F_{k}(i_{k})$ by the universal property of each $\lim F_{k}$ exist an unique morphism $h_{k}:\lim F\rightarrow \lim F_{k}$ and by the universal property of the product $\prod \lim F_{k}$ we have an unique $h:\lim F\rightarrow \prod \lim F_{k}$.

I don't know if i am correct and if i am, how do i prove that $h$ is an isomorphism.

Any idea would help me.

-------------EDIT----------

The morphism $v_{k}^{i}\circ p_{k}:\prod \lim F_{k}\rightarrow F_{k}(i)=F(i)$ for a cone because for $u:i\rightarrow j$ with $i,j\in J_{k}$ for $\lim F_{k}$ we have $v_{k}^{j}=F(u)\circ v_{k}^{i}$ therefore $v_{k}^{j}\circ p_{k}=F(u)\circ v_{k}^{i}\circ p_{k}$.

For $q\circ h=id$ by using the equalities, we have that $v_{i}\circ q\circ h=v_{i}$ by the universal property of $\lim F$ we have $q\circ h=id$.

i don't see how $h\circ q = id$.

• Use \mathoperator{Lim}. – Silvia Ghinassi Nov 14 '15 at 19:04

In order to prove that $h$ is an isomorphism, you have to find a morphism $q$, such that $q=h^{-1}$. Let's find this morphism.
Denote by $p_k\colon\prod\varprojlim F_k\to\varprojlim F_k$ the projection morphism. Let $i\in J$ be an object. Then there exists such $k\in K$, that $i\in J_k$, corresponding morphism $v_k^i\colon\varprojlim F_k\to F_k(i)=F(i)$ and the composition $v_k^i\circ p_k\colon\prod\varprojlim F_k\to F(i)$. By the universal property, we have an arrow $q\colon\prod\varprojlim F_k\to\varprojlim F$, such that for any $i'\in J$ the equality $v_{i'}\circ q=v_{k'}^{i'}\circ p_{k'}$ holds, where $k'\in K$ such that $i'\in J_{k'}$. The equalities $q\circ h=id$ and $h\circ q=id$ also follow from the universal property of limit.
For example, let's prove that $h\circ q=id$. By the universal property of product, it is sufficiently to prove that $p_k\circ h\circ q=p_k$ for every $k\in K$. Let's notice that for every $k\in K$ and $i\in J_k$ the equality $v_k^i\circ p_k=v_k^i\circ p_k\circ h\circ q$ holds, because $v_k^i\circ p_k\circ h=v_i$(by the definition of $h$) and $v_k^i\circ p_k=v_i\circ q$(by the definition of $q$). The universal property of $\varprojlim F_k$ yields $p_k\circ h\circ q=p_k$.
• Thank you!!! in the part $v_{i'}\circ q=v_{k}^{i'}$ i think you missed out $p_{k}$ and how do i see that $h\circ q=id$?. – Liddo Nov 16 '15 at 18:34
• @Liddo Yes, you are right about $p_{k'}$ in the equality - it was my typo. I have added the proof of $h\circ q=id$. – Oskar Nov 16 '15 at 21:08
• Thank you so much!!! i have seen it. $q\circ h=id$ is for the universal property of $\mathop{Lim}F$ and $h\circ q=id$ is for the universal property of each $\mathop{Lim}F_{k}$ (unicity of the morphism)and the universal property of the product (That one is tricky). I am sorry for any trouble. – Liddo Nov 16 '15 at 22:08