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I want to show that proposition$5.33$ in introduction to homological algebra Rotman :let $I$ be a directed set , and let $\{A_i,\alpha_j^i\}$, $\{B_i,\beta_j^i\}$, and $\{C_i,\gamma_j^i\}$ be directed systems of left $R$-modules over $I$ if $r:\{A_i,\alpha_j^i\}\to\{B_i,\beta_j^i\}$ and $s:\{B_i,\beta_j^i\}\to\{C_i,\gamma_j^i\}$ are morphisms of direct systems, and if

$$0\to A_i\xrightarrow{r_i}B_i\xrightarrow{s_i}C_i\to0$$

is exact for each $i\in I$,then there is an exact sequence

$$0\to\varinjlim A_i\xrightarrow{r^\to}\varinjlim B_i\xrightarrow{s^\to }\varinjlim C_i\to0$$

I have same problem to show that ker ${s^\to}\subset$Image$ \ r^\to$. can you help me!thanks.

Bernard
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pink floyd
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2 Answers2

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Take an element $b\in\ker s^\to$, and an element $b_i$ in some $B_i$ such that $\beta_i(b_i)=b $. Then $\gamma_i(s_i(b_i))=0$, so that $\gamma_{ij}(s_i(b_i))=0$ in some $C_j\enspace(j\ge i)$.

As $\,\gamma_{ij}s_i=s_j\beta_{ij}$, this means $\,b_j=\beta_{ij}(b_i)\in\ker s_j$, so there exists $a_j\in A_j$ such that $b_j=r_j(a_j)$. Since $\beta_jr_j=r^\to\alpha_j$, we deduce $$b=\beta_j(b_j)=r^\to(\alpha_j(a_j)),$$ which proves $b$ is the image of $a=\alpha_j(a_j)$ by $r^\to$.

Bernard
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  • @a-lawliet: It is by the construction of the direct limit in the category of sets: it is the disjoint union of the $B_i$s, modulo the equivalence relation $b_i\sim b_j$ iff there exists $k\ge i,j$ such that $\beta_k^i(b_i)=\beta_k^j(b_j)$. – Bernard Oct 28 '19 at 12:59
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Hint: $sr: \{A_i,\alpha^i_j\} \to \{B_i,\beta^i_j\}$ is the zero morphism of direct systems.

Daniel Valenzuela
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