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I am trying to show that projective modules are flat using their defining property that $Hom(P,-)$ is an exact functor when $P$ is projective.

The two ways I know of come down to the fact that projective modules, being summands of free modules which are flat, must also be flat; or, to use derived functors using projective resolutions, and using the long exact sequence of Tor. But I am trying to prove it using the tensor-hom adjunction and the contravariant Yoneda embedding.

My attempt goes like this: supposing that $0\rightarrow A\rightarrow B$ is exact, we want to prove that $0\rightarrow A\otimes P\rightarrow B\otimes P$ is exact when $P$ is projective. We have the following*:

$$ 0\rightarrow A\rightarrow B \text{ exact} \implies Hom(B,C)\rightarrow Hom(A,C)\rightarrow 0 \\ \text { is exact for all C, since Hom(-,C) is contravariant left exact} $$

Now, since $P$ is projective and $Hom(P,-)$ is exact, we also have that this is exact:

$$Hom(P,Hom(B,C))\rightarrow Hom(P,Hom(A,C))\rightarrow 0$$

We can use the tensor-hom adjunction which states that $Hom(P,Hom(-,C))\simeq Hom(P\otimes -, C)$ and get:

$$Hom(B\otimes P, C)\rightarrow Hom(A\otimes P, C) \rightarrow 0 \text{ is exact for all C}$$

Here, I think we have to use the fact that the contravariant Yoneda embedding $y^N=Hom(N,-)$ is fully faithful and preserves exactness, so we must have that $$0\rightarrow A\otimes P\rightarrow B\otimes P \text{ is exact, showing that $P$ is flat}$$

Is this proof correct?

*ADDED: There is a mistake here: as pointed out by Mindlack, this only works when $C$ is injective. Perhaps this modification could fix it: if $A \otimes P$ has an injective hull $I$, take it instead of $C$ - then we still have that $Hom(B\otimes P, I)\rightarrow Hom(A\otimes P, I) \rightarrow 0 $ is exact, which I believe shows that $0\rightarrow A\otimes P\rightarrow B\otimes P $ must be exact: the inclusion of $A\otimes P$ into $I$ lifts to a map from $B\otimes P$ to $I$, and so the kernel of $A\otimes P\rightarrow B\otimes P$ must be 0. So my guess is that if $Hom(Y,I)\rightarrow Hom(X,I)\rightarrow 0$ is exact for all injective $I$, then $0\rightarrow X\rightarrow Y$ must be exact?

B. S.
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  • Your left exactness statement is incorrect, I think: you're basically stating that if $A \subset B$, then any morphism $A \rightarrow C$ extends to a morphism $B \rightarrow C$ for all $C$, ie that all the objects in your category are injective. When a functor (co- or contravariant, no matter) is left exact (which $Hom(-,C)$ is), it means that "short partial" exact sequences mapped to sequences $0 \rightarrow U \rightarrow V \rightarrow W$ are exact. – Aphelli Dec 02 '21 at 14:24
  • you're right! Is there a way to fix this, i.e. by restricting only to injective modules, or something like that? – B. S. Dec 02 '21 at 14:35
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    Yes, there is a fix: just consider instead the exact sequence $0 \rightarrow K \rightarrow X \rightarrow Y$: it means that $K$ has no nonzero morphism to an injective and thus is zero. – Aphelli Dec 02 '21 at 15:34

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