Show that there exist $f_0 : M \to N$ such that the diagram commutes

Question

Let $A$ be a ring, $M,N$ $A$-modules, and consider the following diagram of $A$-modules, where the two horizontal sequences are exact and $f:M\to N$ is $A$-linear:

$$\require{AMScd} \begin{CD} A^{(I_1)} @>>> M @>>> 0 \\ @. @VVfV\\ A^{(J_1)} @>>> N @>>> 0 \end{CD} $$

Let $\epsilon:A^{(I_1)} \to M$ and $\epsilon':A^{(J_1)} \to N$ be the epimorphisms that appear on the diagram, defined by $\epsilon (e_i)=x_i$ where $\{x_i\}_{i \in \hat{I}_1}$ is a system of generators of $M$ and analogously with $N$ and $\epsilon'$.

The question is to show that there exist $f_0:A^{(I_1)} \to A^{(J_1)}$ such that the diagram commutes.

Take $e_i \in A^{(I_1)}$, then $f\circ \epsilon (e_i) = f(x_i) \in N$. Since $\epsilon'$ is an epimorphism, there exist $y \in A^{(J_1)}$ such that $\epsilon'(y)=f(x_i)$. So my idea is to define $f_0$ such that $f_0(e_i)=y$. Now the problem is that there could be several $y \in A^{(J_1)}$ such that $\epsilon'(y)=f(x_i)$.

Question: Is $f_0$ well defined? I'm not being able to understand if it is, or if the dependance of the $y \in A^{(J_1)}$ may affect the good definition.

Answer 1

The notation is quite confusing, so first let's try and simplify it.

You have

1. a module $M$, with set of generators $\{x_i:i\in I\}$,
2. the free module $A^{(I)}$, with basis $\{e_i:i\in I\}$,
3. a module $N$, with set of generators $\{y_j:j\in J\}$,
4. the free module $A^{(J)}$, with basis $\{e'_j:j\in J\}$.
5. a homomorphism $f\colon M\to N$.

There are unique homomorphisms $\varepsilon\colon A^{(I)}\to M$ and $\varepsilon'\colon A^{(J)}\to N$ such that $$ \varepsilon(e_i)=x_i,\quad i\in I \qquad\text{and}\qquad \varepsilon(e'_j)=y_j,\quad j\in J $$ You want to define $f_0\colon A^{(I)}\to A^{(J)}$ such that $$ f\circ\varepsilon=\varepsilon'\circ f_0 $$

In order to define $f_0$ you just have to assign what $f_0(e_i)$ should be, since the domain of $f_0$ is free with basis $\{e_i:i\in I\}$. Now the requirement reads $$ \varepsilon'\circ f_0(e_i)=f\circ \varepsilon(e_i)=f(x_i) $$ For every $i\in I$, choose $z_i\in A^{(J)}$ such that $\varepsilon'(z_i)=f(x_i)$. Then assign $f_0(e_i)=z_i$ and you're done.

Note that this works also if instead of $A^{(J)}\to N$ surjective you take any $K\to N$ surjective.

Even more generally, this works for every diagram with exact rows of the form $$ \begin{CD} P @>\alpha>> M @>>> 0 \\ @. @VVfV \\ K @>\beta>> N @>>> 0 \end{CD} $$ where $P$ is projective. Just use the definition and the diagram $$ \begin{CD} {} @. P \\ @. @VV{f\circ\alpha}V \\ K @>\beta>> N @>>> 0 \end{CD} $$

Answer 2

You can do better than the existence of $y \in A^{J_1}$ such that $\epsilon'(y)=f(x_i)$. Because $f(x_i)=y_j$ for $\{y_j\}_{j\in \hat J_1}$ a system of generators on $N$ we have $\epsilon'( e'_j) = y_j = f(x_i)$. And define $f_0(e_i) = e'j$.

Answer 3

This is a rather odd question: the question in the title is quite different to that in the text.

For the one in the text, there is the basic fact ("free modules are projective") that if $F$ is a free module, $\phi:N'\to N$ is surjective and $f\in\text{Hom}_A(F,N)$ then $f$ lifts to $g\in\text{Hom}_A(F,N')$, that is there is $g$ with $f=\phi\circ g$. Here you take $F=A^{(I_1)}$, $\phi$ the given map from $A^{(I_2)}$ to $N$, and $f$ the composite map $A^{(I_1)}\to M\to N$.

One proves this by noting that maps from a free module $A^{(I_1)}$ to $N$ are determined freely by the images of the standard generators of $A^{(I_1)}$ in $N$. Each of these can be lifted to $N'$ giving rise to a map from $A^{(I_1)}$ to $N'$.

For the question in the title, if $M$ is finitely generated projective, then $M\oplus M'\cong A^n$ for some module $M'$ and some integer $n$. Then $\text{Hom}_A(M,A)\oplus\text{Hom}_A(M',A)\cong \text{Hom}_A(A^n,A)$ etc.