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.


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 $h\circ q = id$.

  • 1
    $\begingroup$ Use \mathoperator{Lim}. $\endgroup$ – 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 $p_k\circ h\circ q=p_k$.

  • $\begingroup$ 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 $h\circ q=id$?. $\endgroup$ – Liddo Nov 16 '15 at 18:34
  • $\begingroup$ @Liddo Yes, you are right about $p_{k'}$ in the equality - it was my typo. I have added the proof of $h\circ q=id$. $\endgroup$ – Oskar Nov 16 '15 at 21:08
  • $\begingroup$ 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. $\endgroup$ – Liddo Nov 16 '15 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.