Ok, this problem is driving me nuts. At first, I thought I did it. But when reading another textbook (having a similar proposition, they (the problem in my textbook, and the proposition in the other book) are not exactly the same), so I went over my work again, and figured out a big flaw in my work, so I redid it, but I just couldnot complete it. Here is the problem of my textbook:


Notice: In the following problem $A$; $J$, and $P$ are all $R-$modules.

Prove that $J$ is injective iff for every epimorphism $f:A \to J$, and monomorphism $g:A \to P$ (where $P$ is some projective module), there must exist a homomorphism $\varphi: P \to J$, such that $\varphi g = f$.

And the other version is without the epimorphism part, it just requires $f$ to be any module map.

And here's my work

  1. The $\Rightarrow:$ part should be straight-forward from the definition of injective modules.
  2. Now, the $\Leftarrow:$ part. I'll prove $J$ injective, by proving that every short exact sequence of the form $0 \to J \xrightarrow{\chi} A \xrightarrow{\sigma} B \to 0$ splits. And to prove that, my aim is to construct an inverse of $\chi$. So consider the following diagram:

enter image description here

Where $R^A$, and $R^J$ are respectively free modules generated by elements of $A$, and $J$. $i$ is the natural injection, since $J$ can be thought of as a submodule of $A$, and $p_1$, $p_2$ are epimorphisms.

Since $R^A$ is free, hence projective. So according to the problem, there exists $\gamma: R^A \to J$, so that the bottom left triangle commutes.

At first, I thought I can construct $\beta: A \to J$ by using the on to property of $p_2$, i.e, for every $a \in A$, there exists $(\alpha_i)_{i \in A} \in R^A$, such that $p_2((\alpha_i)_{i \in A}) = a$. And I can define $\beta (a) = \gamma((\alpha_i)_{i \in A})$. However, with that definition, I cannot prove that $\beta$ is well-defined, let alone, a module map, and I don't really think it's a map at all. But, maybe I'm wrong.

So, can someone help me, am I missing something here? Or is the book wrong? Or should I start with a different way?

Thanks everyone,

And have a good day,

  • 3
    $\begingroup$ Your proof cannot work since $\gamma$ exists without any assumption on $J$. Extension on free modules is trivial. By the way, you should write somewhere in the beginning that you work with $R$-modules. $\endgroup$ Apr 25, 2013 at 10:10
  • 1
    $\begingroup$ When you don't require that $f$ is epi, you may even assume that $P=R$. This is Baer's criterion for injectivity. $\endgroup$ Apr 25, 2013 at 10:17
  • $\begingroup$ Hint: Let $B$ be the pull-back of $p_1$ and $\chi$. The pull-back square gives a surjection $B \to J$ and an injection $B \to R^A$. Apply the assumed property of $J$ to these maps to find $\varphi \colon R^A \to J$. Then use that the square involving $P$ is also a push-out to find a splitting of $\chi$ using $\varphi$. $\endgroup$
    – Martin
    Apr 25, 2013 at 10:31
  • $\begingroup$ I get the first 2 lines, but I'm a little bit lost on the last line, what's $P$? $\endgroup$
    – user49685
    Apr 25, 2013 at 12:47
  • 1
    $\begingroup$ No, I meant to say that the pull-back $$\begin{array}{lll} J & \xrightarrow{\chi} & A \\ \uparrow & & \uparrow \\ B & \rightarrow & R^A \end{array}$$ is both is also a push-out ($P$ was a typo for $B$). Now use that $1_J \colon J \to J$ and $\varphi\colon R^A \to J$ give a commutative square involving $B, J, R^A$ and the map $A \to J$ obtained from the push-out yields a splitting of $\chi$. $\endgroup$
    – Martin
    Apr 25, 2013 at 21:53

2 Answers 2


I think it could be easier to work directly with the basic definition of injectivity, i.e. that if $A \hookrightarrow B$ is an embedding, then any $A \to J$ extends to $B$.

Now you are supposed to reduce to the case when $A \to J$ is surjective, and $B$ is projective. You can always just add a copy of $J$ to $A$ and $B$ to get surjectivity. And you can always write $B \oplus J$ as a quotient of some projective $P$ and form an appropriate pull-back to assume $B$ is projective. I leave the details to you.

  • $\begingroup$ I don't really get the last part, but I'll give it a try. So basically, what I'm trying to prove is, given any injection $i: I \to R$, where $I$ is an ideal of $R$, and any homomorphism $f: I \to J$, it can be lifted. If $f$ this is epi, then everything works out as expected, if not, I'll consider the map: $\tilde{f}: I \oplus J \to J$, where $\tilde{f}(i; j) = f(i) + j$, this map is indeed epi, and the map $f \oplus \mbox{id}_J : I \oplus J \to R \oplus J$ is an injection. So what's left is to prove $R \oplus J$ is projective. $\endgroup$
    – user49685
    Apr 25, 2013 at 13:19
  • $\begingroup$ My comment is too long, so I'll split it in two: I know how to form pull-backs, but I don't see how pull-backs and projectivity relate? So is what I do above the same as what you are hinting me? And hopefully, I'm still on track, ain't I? @.@ $\endgroup$
    – user49685
    Apr 25, 2013 at 13:21
  • $\begingroup$ Ok, on thinking a bit, it doesn't seem true at all, if, say $R \oplus J$ is projective, then $J$ must be projective. And clearly $J$ is not projective. $\endgroup$
    – user49685
    Apr 25, 2013 at 13:28
  • $\begingroup$ @user49685: Dear user, As I indicated in my sketch, there is another step needed: you need to choose a projective module that surjects onto $R \oplus J$, and then make a corresponding modification of $I \oplus J$. Regards, $\endgroup$
    – Matt E
    Apr 25, 2013 at 13:52
  • $\begingroup$ I'm sorry, but I still don't see where it goes, don't be mad at me... Ok, so the projective module that surjects onto $R \oplus J$ is clearly $R^{R \oplus J}$, and $I \oplus J$ should become $R^{I \oplus J}$, the maps are still injective, and surjective, so there exists $\gamma: R^{R \oplus J} \to J$, how can I go from $\gamma$ to some map $\beta: R \to J$, isn't it still where I'm stuck? $\endgroup$
    – user49685
    Apr 25, 2013 at 14:18

Ok, so I've been working on this problem for more than half a day, and finally it pays off. I don't really know if there's any flaws left in this solution, but it seems true to me. It's midnight here, and I may make some kind of stupid mistake. So if someone can confirm it for me, I would be very glad.

My aim is to use Baer's criterion to solve it. So for every injective $i: I \to R$, and for every homomorphism $f: I \to J$, I need to lift it to some $g: R \to J$.

  1. If $f$ is epi, then it's done, since $R$ is free, hence projective.
  2. If $f$ is not epi, I construct an epimorphism $\gamma: R^J \to J$. So we have the following diagram: $\begin{array}{ccc} I \oplus R^J & \xrightarrow{i\oplus\mbox{id}_{R^J}} & R \oplus R^J \\ \downarrow^{\tilde{f}} \\ J \end{array}$

where $\tilde{f}(i; (r_k)_{k \in J}) = f(i) + \gamma((r_k)_{k \in J})$, since $\tilde{f}$ is epi, and $i \oplus \mbox{id}_{R^J}$ is injective, and $R \oplus R^J$ is free hence projective, so there's a homomorphism $\varphi: R \oplus R^J \to J$, such that the diagram commutes. It's easy to see that $g = \varphi \circ j$ is the module map I'm looking for, where $j$ is an injective map from $R$ to $R \oplus R^J$, i.e $j(b) = (b; 0_{R^J})$.

Does this look good?

  • 1
    $\begingroup$ Dear user, This looks good. And I'm sorry if my partial answer caused frustation; hopefully working out the answer for yourself made it worthwhile, though. Best wishes, $\endgroup$
    – Matt E
    Apr 26, 2013 at 10:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .