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In page 33 of Pierre Schapira's Algebra and Topology, it is briefly mentioned that given functor $\alpha:I\rightarrow C$, if $i_0$ is initial in $I$, then there is an isomorphism

$\lim_\longleftarrow\alpha \approx \alpha(i_0)$

To try to verify this, I try the following.

First denote $\{\lambda_i\}_i$ as the morphisms associated with limit $L$. For given $i\in I$, associate $\rho_{i}$ as the application of $\alpha$ to the unique morphism from $e_{0}$ to $i$. Then we have by the definition of the limit, a morphism $\mu:\alpha(i_0)\rightarrow L$ such that $\lambda(0)\circ\mu=id$.

How should I prove $\mu\circ\lambda(0)=id$? (if this is the right way to approach the problem)

Arnaud D.
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rr01
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1 Answers1

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Let $\alpha:f_{jm}:\alpha(i_j)\to \alpha(i_m)$ be a diagram in $C.$ The claim is that $\alpha(i_0)$ is a limit for $\alpha$, which means we need to find a limit cone, and then verify the universal property of limit. These facts follow from

$1).\ $ Since $i_0$ is initial in $I$ it makes sense to define $\tau_{j}=\alpha(i_0\to i_j)$, and then $\tau_m=f_{jm}\circ \tau_j$ which means that $(\tau_j)_{j\in I}$ is a cone over $\alpha$.

$2).$ Let $(x\in C,(\lambda_j)_{j\in I})$ be a cone over $\alpha$. We need a unique arrow $\varphi:x\to \alpha(i_0)$ such that $\tau_m\circ \varphi=\lambda_m.$ If we take $\varphi$ to be the distinguished arrow, $\lambda_{i_0}$ then $\tau_m\circ \lambda_{i_0}=f_{{i_0}m}\circ \tau_{i_0}\circ \lambda_{i_0}=f_{{i_0}m}\circ \lambda_{i_0}=\lambda_m.$

$3).\ $ It is clear that $\varphi $ is uniquely determined, and this concludes the proof.

Matematleta
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