## FANDOM

36 Pages

AMS GSM 104

Errata available online: [1]

"This text presents an introduction to algebra suitable for upper-level undergraduate or beginning graduate courses. While there is a very extensive oﬀering of textbooks at this level, in my experience teaching this material I have invariably felt the need for a self-contained text that would start ‘from zero’ (in the sense of not assuming that the reader has had substantial previous exposure to the subject), but impart from the very beginning a rather modern, categorically-minded viewpoint, and aim at reaching a good level of depth. Many textbooks in algebra satisfy brilliantly some, but not all of these requirements. This book is my attempt at providing a working alternative."

# Chapter IX : Homological algebraEdit

## Section IX.1 (Un)necessary categorical preliminaries Edit

• After Example 1.11: The coproduct $A \sqcup B$ is endowed with morphisms $i_A: A \to A \sqcup B$, $i_B: B \to A \sqcup B$ which are easily checked to be monomorphisms (do this!): We have $\pi_A \circ i_A = 1_A$ by definition, or in other words $\pi_A$ is a left-inverse of $i_A$. Since monomorphisms are defined by cancellation, this is actually even a bit stronger than being a monomorphism (roughly the equivalent of "unit implies non-zerodivisor" in a commutative ring.)
• Proposition 1.12, $\pi_B$ is the cokernel of $i_A$: The proof given here is bizarrely stated. I had to work things out on my own, after which I eventually realized that my method matched the books. Don't, as suggested, stare at the diagram; the diagram is weird and unhelpful. Here's what it says: We already know that $\pi_B \circ i_A = 0$, so what remains to show is that for any object C and any map $\gamma : A \sqcup B \to C$ with $\gamma \circ i_A = 0$, there exists a unique map $\delta : B \to C$ such that $\delta \circ \pi_B = \gamma$. In particular, we claim that $\delta = \gamma \circ i_B$. To see this, consider the (unique!) map $g : = (0 \sqcup (\gamma \circ i_B)) : A \sqcup B \to C$; that is, the unique map such that $g \circ i_A = 0$ and $g \circ i_B = \gamma \circ i_B$. By definition, $\gamma$ satisfies both equations, so $g = \gamma$. On the other hand, we can also check that $(\gamma \circ i_B \circ \pi_B)$ satisfies both equations, so $g = \gamma \circ i_B \circ \pi_B$. Therefore $\gamma = \gamma \circ i_B \circ \pi_B = \delta \circ \pi_B$, which is what we wanted to show.
• Exercise 1.14: The word 'sheaf' should be replaced by 'presheaf' here.