Q:

Is the canonical map $\pi: F \to R$ of an algebra $R$ over a field $F$ injective if and only if $R$, as a ring, is not the zero ring?

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).
I refer to Section B.1, Section B.3 (part 1) and Section B.3 (part 2) and to Invariant Polynomials on $\mathfrak{gl} (r,F)$.

Question: Is the proof of Proposition B.5 actually assuming $R$ is not the algebra with exactly 1 element unless there's some convention like "with identity" means "not the zero ring"?

I think the statement Proposition B.5 need not assume $R$ is not the algebra with exactly 1 element, but a part of the proof does, namely that the canonical map $\pi: F \to R$ is injective if and only if $R$ is not the algebra with exactly 1 element. I believe the preceding sentence is equivalent to Wikipedia's "$A$ is the zero ring, or that $\eta$ is injective." I of course assume $R$, as a ring, is the zero ring if and only if $R$, as an algebra, is the algebra with exactly 1 element. I notice another textbook made the same error, so there might be a convention I'm not aware of.

Update: I notice it says "with identity 1" and not just "with identity" or something like "unital". Probably a convention.

A:

The statement that $f\mapsto f\cdot 1$ is injective is indeed wrong since $R$ could be the zero algebra.  Note though that the injectivity of this map is never used in the proof so the proof remains correct without it.

