Lemma 13.5.2. Let $\mathcal{D}$ be a pre-triangulated category. Let $S \subset \text{Arrows}(\mathcal{D})$.
Proof. Assume $S$ contains all identities and MS6 holds. Let $f : X \to Y$ be an isomorphism of $\mathcal{D}$. Consider the diagram
\[ \xymatrix{ 0 \ar[r] \ar[d]^1 & X \ar[r]_1 \ar[d]^1 & X \ar[r] \ar@{-->}[d] & 0[1] \ar[d]^{1[1]} \\ 0 \ar[r] & X \ar[r]^ f & Y \ar[r] & 0[1] } \]
The rows are distinguished triangles by Lemma 13.4.9. By MS6 we see that the dotted arrow exists and is in $S$, so $f$ is in $S$.
Assume MS1, MS5, MS6. Suppose that $f : X \to Y$ is a morphism of $\mathcal{D}$ and $t : X \to X'$ an element of $S$. Choose a distinguished triangle $(X, Y, Z, f, g, h)$. Next, choose a distinguished triangle $(X', Y', Z, f', g', t[1] \circ h)$ (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram
\[ \xymatrix{ X \ar[r] \ar[d]^ t & Y \ar[r] \ar@{..>}[d]^{s'} & Z \ar[r] \ar[d]^1 & X[1] \ar[d]^{t[1]} \\ X' \ar[r] & Y' \ar[r] & Z \ar[r] & X'[1] } \]
with moreover $s' \in S$. This proves LMS2. The proof of RMS2 is dual. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: