Section 3. (0MVE) · ScalingStacks

3. Existence of an $S^{\perp_{\infty}}$ -preenvelope

In order to obtain an $S^{\perp_{\infty}}$ -preenvelope, we first show how to construct an $S^{\perp_{n}}$ -preenvelope for each $n\in\mathbb{N}$ . It is useful to refer to a simply connected $n$ -corigid object of a triangulated category:

0MUF

Definition 3.1. Let $S$ be an object of $\mathcal{T}$ . We shall call $S$ a simply connected $n$ -corigid object of $\mathcal{T}$ if it satisfies the following assumptions:

$(1)$ $S$ is $n$ -corigid, that is, $\textnormal{Hom}(\Sigma^{i}S,S)=0$ for $0<i<n$ ;

$(2)$ $\textnormal{Hom}(S,\Sigma S)=0$ ;

$(3)$ $\textnormal{End}(S)$ is a division ring.

0MUG

Proposition 3.2. Let $\mathcal{T}$ be a triangulated category with set indexed coproducts. Suppose $S$ is a compact simply connected $(n+1)$ -corigid object of $\mathcal{T}$ . Then, for each object $M$ of $\mathcal{T}$ there exists an $S^{\perp_{n}}$ -preenvelope $\mu:M\rightarrow\bar{M}$ .

0MUH

Proof: Let $M$ be an arbitrary object of $\mathcal{T}$ . We first construct a chain of objects and morphisms,

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{n-1}}}{{\longrightarrow}}M_{n}=\bar{M}

with $M_{k}\in S^{\perp_{k}}$ for each $k\geqslant 1$ , inductively using distinguished triangles. Secondly, we verify that the composite of these maps is an $S^{\perp_{n}}$ -preenvelope.

Write $M=M_{0}$ . Let $n=1$ ; we construct an object $M_{1}$ and a morphism $\mu_{0}:M_{0}\rightarrow M_{1}$ such that $\textnormal{Hom}(S,\Sigma M_{1})=0$ . If $\textnormal{Hom}(S,\Sigma M_{0})=0$ then set $M_{1}=M_{0}$ and $\mu_{0}=1_{M_{0}}$ , the identity map on $M_{0}$ . If not, we can choose a, possibly infinite, coproduct $S^{(m_{1})}$ of copies of $S$ and a nonzero morphism $S^{(m_{1})}\rightarrow\Sigma M_{0}$ which becomes a surjection under the functor $\textnormal{Hom}(S,-)$ . Since the endomorphism ring $\textnormal{End}(S)$ is a division ring we can, moreover, choose $m_{1}$ so that this morphism becomes an isomorphism under $\textnormal{Hom}(S,-)$ . We now extend this morphism to a distinguished triangle:

(3.1)

S^{(m_{1})}\rightarrow\Sigma M_{0}\rightarrow\Sigma M_{1}\rightarrow\Sigma S^{% (m_{1})}.

Applying $\textnormal{Hom}(S,-)$ to (3.1) gives the exact sequence:

\textnormal{Hom}(S,S^{(m_{1})})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}% }\textnormal{Hom}(S,\Sigma M_{0})\rightarrow\textnormal{Hom}(S,\Sigma M_{1})% \rightarrow\textnormal{Hom}(S,\Sigma S^{(m_{1})}).

Since $\textnormal{Hom}(S,\Sigma S^{(m_{1})})=0$ , we get $\textnormal{Hom}(S,\Sigma M_{1})=0$ .

Now suppose $k\geqslant 1$ and suppose we have constructed a chain of objects and morphisms

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{k-1}}}{{\longrightarrow}}M_{k}

with $M_{i}\in S^{\perp_{i}}$ for $1\leqslant i\leqslant k$ , and where $\mu_{i}:M_{i}\rightarrow M_{i+1}$ is either the identity map or sits in a distinguished triangle

\Sigma^{-(i+1)}S^{(m_{i+1})}\rightarrow M_{i}\stackrel{{\scriptstyle\mu_{i}}}{% {\longrightarrow}}M_{i+1}\rightarrow\Sigma^{-i}S^{(m_{i+1})}.

If $\textnormal{Hom}(S,\Sigma^{k+1}M_{k+1})=0$ then set $M_{k+1}=M_{k}$ and take $\mu_{k}:M_{k}\rightarrow M_{k+1}$ to be the identity map $1_{M_{k}}$ . If not, we can choose a, possibly infinite, coproduct $S^{(m_{k+1})}$ of copies of $S$ and a nonzero morphism $S^{(m_{k+1})}\rightarrow\Sigma^{k+1}M_{k}$ which becomes an isomorphism under $\textnormal{Hom}(S,-)$ , and then extend it to a distinguished triangle:

(3.2)

S^{(m_{k+1})}\rightarrow\Sigma^{k+1}M_{k}\rightarrow\Sigma^{k+1}M_{k+1}% \rightarrow\Sigma S^{(m_{k+1})}.

As above, an argument from the long exact sequence of Hom-sets arising from (3.2) shows that

\textnormal{Hom}(S,\Sigma^{i}M_{k+1})=0\textnormal{ for }i=1,\ldots,k+1.

The case $i=k$ follows by the injectivity of $\textnormal{Hom}(S,S^{(m_{k+1})})\stackrel{{\scriptstyle\sim}}{{% \longrightarrow}}\textnormal{Hom}(S,\Sigma^{k+1}M_{k})$ ; and the case $i=k+1$ by its surjectivity.

Hence, inductively we obtain a chain of objects and morphisms of $\mathcal{T}$ ,

(3.3)

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{n-1}}}{{\longrightarrow}}M_{n},

where each map $\mu_{k}:M_{k}\rightarrow M_{k+1}$ is either the identity map or sits in a distinguished triangle

(3.4)

\Sigma^{-(k+1)}S^{(m_{k+1})}\rightarrow M_{k}\stackrel{{\scriptstyle\mu_{k}}}{% {\longrightarrow}}M_{k+1}\rightarrow\Sigma^{-k}S^{(m_{k+1})}.

To see that the composite $\mu=\mu_{n-1}\circ\cdots\circ\mu_{1}\circ\mu_{0}$ from (3.3) is an $S^{\perp_{n}}$ -preenvelope, we shall show that for each $X\in S^{\perp_{n}}$ the map $\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)$ induced by $\mu_{k}$ is a surjection. Without loss of generality we may assume that each map $\mu_{k}$ sits in a distinguished triangle (3.4) above, because if $\mu_{k}=1_{M_{k}}$ , then the map $\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)$ is trivially an isomorphism for all $X\in\mathcal{T}$ .

Let $X\in S^{\perp_{n}}$ ; applying $\textnormal{Hom}(-,X)$ to distinguished triangle (3.4), we get the long exact sequence of Hom-sets below:

(\Sigma^{-k}S^{(m_{k+1})},X)\rightarrow(M_{k+1},X)\rightarrow(M_{k},X)% \rightarrow(\Sigma^{-(k+1)}S^{(m_{k+1})},X),

where we have written $(A,B)$ as a shorthand for $\textnormal{Hom}(A,B)$ . Now since we have $\textnormal{Hom}(S^{(m_{k+1})},\Sigma^{k}X)=\textnormal{Hom}(S^{(m_{k+1})},% \Sigma^{(k+1)}X)=0$ for $k=1,\ldots,n-1$ , the map

\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)

induced by $\mu_{k}$ is an isomorphism for $k=1,\ldots,n-1$ and a surjection for $k=0$ . Hence, writing $\bar{M}=M_{n}$ , the composite $\mu:M\rightarrow\bar{M}$ is an $S^{\perp_{n}}$ -preenvelope. $\Box$

0MUI

Lemma 3.3. Suppose further that $S$ is a simply connected corigid object of $\mathcal{T}$ . Then, for the $S^{\perp_{n}}$ -preenvelope, $\mu:M\rightarrow\bar{M}$ , obtained in Proposition 3.2, we have that

\textnormal{Hom}(S,\Sigma^{i}\mu):\textnormal{Hom}(S,\Sigma^{i}M)\rightarrow% \textnormal{Hom}(S,\Sigma^{i}\bar{M})

is an isomorphism for all $i<1$ .

0MUJ

Proof: Applying the functor $\textnormal{Hom}(S,-)$ to distinguished triangle (3.4),

\Sigma^{-(k+1)}S^{(m_{k+1})}\rightarrow M_{k}\stackrel{{\scriptstyle\mu_{k}}}{% {\longrightarrow}}M_{k+1}\rightarrow\Sigma^{-k}S^{(m_{k+1})},

for $0\leqslant k<n$ in the proof of Proposition 3.2 shows that

\textnormal{Hom}(S,\Sigma^{i}\mu):\textnormal{Hom}(S,\Sigma^{i}M_{k})% \rightarrow\textnormal{Hom}(S,\Sigma^{i}M_{k+1})

is an isomorphism for all $i<k+1$ . The isomorphism for $i=k$ follows by the fact that the morphism $S^{(m_{k+1})}\rightarrow\Sigma^{k+1}M_{k}$ in (3.4) is constructed to be an isomorphism under $\textnormal{Hom}(S,-)$ . Hence the composite $\mu=\mu_{n-1}\circ\cdots\circ\mu_{1}\circ\mu_{0}$ is an isomorphism under $\textnormal{Hom}(S,\Sigma^{i}-)$ for all $i<1$ . $\Box$

In order to obtain an $S^{\perp_{\infty}}$ -preenvelope we need to introduce the key tool, which is called the homotopy colimit. The following definition is taken from [15].

0MUK

Definition 3.4. Let $\mathcal{T}$ be a triangulated category with set indexed coproducts. Let

X_{0}\stackrel{{\scriptstyle f_{0}}}{{\longrightarrow}}X_{1}\stackrel{{% \scriptstyle f_{1}}}{{\longrightarrow}}X_{2}\stackrel{{\scriptstyle f_{2}}}{{% \longrightarrow}}X_{3}\stackrel{{\scriptstyle f_{3}}}{{\longrightarrow}}\cdots

be a sequence of objects and morphisms in $\mathcal{T}$ . The homotopy colimit $\textnormal{hocolim}(X_{i})$ is constructed by extending the map

\coprod_{i=0}^{\infty}X_{i}\stackrel{{\scriptstyle 1-\rm{shift}}}{{% \longrightarrow}}\coprod_{i=0}^{\infty}X_{i}

to a distinguished triangle:

\coprod_{i=0}^{\infty}X_{i}\stackrel{{\scriptstyle 1-\rm{shift}}}{{% \longrightarrow}}\coprod_{i=0}^{\infty}X_{i}\longrightarrow\textnormal{hocolim% }(X_{i})\longrightarrow\Sigma\coprod_{i=0}^{\infty}X_{i}{}{}.

We will need the following lemma.

0MUL

Lemma 3.5 ([15], Lemma 2.8). Suppose $S$ is a compact object of a triangulated category $\mathcal{T}$ and we have a sequence of objects and morphisms of $\mathcal{T}$ :

X_{0}\rightarrow X_{1}\rightarrow X_{2}\rightarrow X_{3}\rightarrow\cdots

then $\textnormal{colim}\,(\textnormal{Hom}(S,X_{n}))\cong\textnormal{Hom}(S,% \textnormal{hocolim}X_{n})$ .

0MUM

Proposition 3.6. Let $\mathcal{T}$ be a triangulated category with set indexed coproducts. Suppose $S$ is a compact simply connected corigid object of $\mathcal{T}$ . Then, for each object $M$ of $\mathcal{T}$ there exists an $S^{\perp_{\infty}}$ -preenvelope $\mu:M\rightarrow\bar{M}$ .

0MUN

Proof: Let $M$ be an object of $\mathcal{T}$ and write $M=M_{0}$ . Let $X\in S^{\perp_{\infty}}$ and suppose we have a morphism $\alpha_{0}:M_{0}\rightarrow X$ . By the argument of Proposition 3.2 we can construct the following commutative diagram:

with $M_{n}\in S^{\perp_{n}}$ for each $n\geqslant 1$ . We now construct the homotopy colimit, $\textnormal{hocolim}(M_{i})$ . By construction, the composite

\coprod_{i=0}^{\infty}M_{i}\stackrel{{\scriptstyle 1-\textnormal{shift}}}{{% \longrightarrow}}\coprod_{i=0}^{\infty}M_{i}\stackrel{{\scriptstyle\langle% \alpha_{i}\rangle}}{{\longrightarrow}}X

is zero, so that we have the following commutative diagram:

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That is, every morphism $M\rightarrow X$ factors through $\textnormal{hocolim}(M_{i})\rightarrow X$ .

Now we have:

$\displaystyle\textnormal{Hom}(S,\Sigma^{j}\textnormal{hocolim}(M_{i}))$	$\displaystyle\cong$	$\displaystyle\textnormal{Hom}(S,\textnormal{hocolim}(\Sigma^{j}M_{i}))$
	$\displaystyle\cong$	$\displaystyle\textnormal{colim}\ \textnormal{Hom}(S,\Sigma^{j}M_{i})$
	$\displaystyle=$	$\displaystyle 0$

for $j\geqslant 1$ . We obtain the first isomorphism because the homotopy colimit commutes with the suspension functor and the second isomorphism by Lemma 3.5. The final equality is a consequence of the fact that $\textnormal{Hom}(S,\Sigma^{j}M_{i})=0$ for $i$ sufficiently large and $j\geqslant 1$ . Hence we have $\textnormal{hocolim}(M_{i})\in S^{\perp_{\infty}}$ . Therefore, setting $\bar{M}=\textnormal{hocolim}(M_{i})$ , we obtain an $S^{\perp_{\infty}}$ -preenvelope $\mu:M\rightarrow\bar{M}$ . $\Box$

David Pauksztello

Original source: arXiv:0705.0102v2

3. Existence of an S⟂∞-preenvelope

3. Existence of an $S^{\perp_{\infty}}$ -preenvelope