2 out of 3 property for faithfully flat ring maps

Proposition 1: Let $f : A\to B$ and $g : B\to C$ be ring maps such that $g$ is faithfully flat. Then the composition $gf$ is flat (resp. faithfully flat) if and only if $f$ is flat (resp. faithfully flat).

Proof: Certainly flatness (resp. faithful flatness) of $f$ implies flatness (resp. faithful flatness) of $gf.$

Conversely, suppose that $gf$ is flat, and let $M'\to M\to M''$ be an exact sequence of $A$-modules. Then flatness of $gf$ implies that $M'\otimes_A C\to M\otimes_A C\to M''\otimes_A C$ is exact as well, whence by faithful flatness of $g$ we have that $M'\otimes_A B\to M\otimes_A B\to M''\otimes_A B$ is exact.

If $gf$ is faithfully flat, and $M'\to M\to M''$ is a sequence of $A$-modules such that the composition $M'\to M''$ is $0,$ then faithful flatness of $gf$ implies that $M'\to M\to M''$ is exact if and only if $M'\otimes_A C\to M\otimes_A C\to M''\otimes_A C$ is. Faithful flatness of $g$ implies that this is exact if and only if $M'\otimes_A B\to M\otimes_A B\to M''\otimes_A B$ is. $\square$

This result has been surprisingly hard to track down in the literature, despite the simplicity of the proof. (The proof and statement of proposition 1 in the flat case can be found here; I do not know of anywhere that the version of proposition 1 for faithful flatness exists, although I'm sure I just haven't searched carefully enough.) Note that this proposition is not true only assuming flatness of $g,$ as evidenced by the composition $k[t^2, t^3]\to k[t]\to k(t).$

I was originally interested in the situation where we assume that the composition is flat, and use that to deduce that $f$ is flat. However, I was wondering whether the following stronger proposition is true.

Proposition 2: Let $f : A\to B$ and $g : B\to C$ be ring maps such that the composition $gf : A\to C$ is faithfully flat. Then $f$ is faithfully flat.

Disclosure: I asked this question in the hopes that this 2 out of 3 property for faithfully flat morphisms will more easily searchable for anyone trying to find a result of this type in the future.

Edit: Proposition 2 is in fact false! The original answer is below in spoiler tags, but SashaP has pointed out a counterexample: if $f : k[x]\to k[x]\times k$ is the map induced by the identity on $k[x]$ and $x\mapsto 0$ from $k[x]\to k,$ and $g : k[x]\times k\to k[x]$ is the projection onto the first factor, then $gf = \operatorname{id}_{k[x]},$ but $f$ is not faithfully flat.

Notice that $B = k[x]\times k\cong k[x]\oplus k,$ and $k$ is not flat over $k[x].$ In particular, we have an exact sequence of $k[x]$-modules $$ 0\to k[x]\xrightarrow{\cdot x} k[x]\xrightarrow{\pi} k[x]/(x)\to 0 $$ and upon tensoring with $B,$ we get $$ k[x]\oplus k\xrightarrow{\cdot x\oplus 0}k[x]\oplus k\xrightarrow{\pi\oplus\operatorname{id}_k}k[x]/(x)\oplus k\to 0, $$ which is no longer exact on the left.

The error in the proof is that it is not true that $A\to B$ is faithfully flat if and only if for every $A$-module $N,$ the canonical map $N\to N\otimes_A B$ is injective. We must also suppose that the map $A\to B$ is flat in order to get the equivalence. Indeed, $k[x]\to k[x]\times k$ (or more generally, $A\to B\times B'$ where $B$ is faithfully flat over $A$ and $B'$ is not flat over $A$) will provide a counterexample to the original claim.

Indeed, Proposition 2 is true.

Proof: Recall that $A\to B$ is faithfully flat if and only if the canonical map $N\to N\otimes_A B$ is injective for every $A$-module $N.$ To that end, let $N$ be an $A$-module. We need to prove that $N\to N\otimes_A B$ is injective, but we know by faithful flatness of $gf$ that the composition $N\to N\otimes_A B\to N\otimes_A C$ is injective. Thus, it follows that $N\to N\otimes_A B$ must be injective. $\square$

Certainly, it need not be true that $B\to C$ be faithfully flat even if both $A\to B$ and $A\to C$ are (consider the composition $k\to k[X]\to k(X),$ for $k$ a field). So the best we can hope for is Proposition 2.