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Theorem

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Question

My only question is: why is the diagrammatic condition equivalent to the Remark? I'm hoping to see a rigorous proof of this claim.

IsaacR24
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1 Answers1

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We are dealing with arbitrary f.g. free $F_1,F_0$ and with arbitrary $n,m$. Fix $n$ and $m$ now and declare $F_1=R^n$ and $F_0=R^m$.

Maps $F_1\to F_0$ are given precisely by matrices $[r_{\bullet,\bullet}]$. Maps $F_1\to B'$ are given precisely by tuples $(b'_1,\cdots,b'_n)$ and $F_0\to B$ precisely by tuples $(b_1,\cdots,b_m)$. Work with the standard ordered bases. The diagram commutes exactly when it commutes when restricted to every basis element of $F_1$. "By linear algebra", we see the diagram commutes exactly when: $$\tag{$\ast$}\lambda b'_j=\sum_{i=1}^mr_{j,i}b_i$$Holds for all $1\le j\le n$. The $j$th basis element of $F_1$ is mapped to $\lambda b'_j$ by definition, in the lower route. In the upper route, it is mapped first to the $j$th column of the matrix, $(r_{j,1},r_{j,2},\cdots,r_{j,m})$ and then mapped, by definition, to $r_{j,1}b_1+r_{j,2}b_2+\cdots$, hence $(\ast)$.

Diagonal lifts $F_0\to B'$ making the upper triangle commute are precisely given by tuples $(h_1,\cdots,h_m)$ such that $(\ast)$ holds (without a $\lambda$) with $h_i$ in the stead of $b_i$ for each $i$, for the same reasons.

FShrike
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  • Thanks this fully addresses my question! For the lower triangle to commute, I believe this is equivalent to saying that $ \lambda h_j = b_j $ for each $j$. – IsaacR24 Nov 24 '23 at 13:54
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    @IsaacR24 Yes, that's right. You'll come across (or already have come across) something similar to this if you dig into the equational characterisation of flatness of a module; the fact that submodules are pure iff. they meet this equational characterisation is very related – FShrike Nov 24 '23 at 14:14