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In Theorems 4.7 and 8.4 Hilton & Stammbach give two lists of 5 different characterizations of projective and injective modules, respectively. Even though I can follow the proofs they give, I'd like to get rid of the characterization involving free and cofree modules (the fifth one). Why? Because I don't want to go through the trouble of defining cofree modules (please don't insist on this).

The theorems say that for a module $P$ the following properties are equivalent:

1) $P$ is projective.

2) The functor $\operatorname{Hom}_\Lambda(P,\,\cdot\,)$ is right exact.

3) For every epimorphism $\epsilon:B\to P$, there exists a morphism $\sigma:P\to B$ such that $\epsilon\sigma=1_P$.

4) $P$ is a direct summand in every module of which it is a quotient.

and that for a module $I$ the following properties are equivalent:

1') $I$ is injective.

2') The functor $\operatorname{Hom}_\Lambda(\,\cdot\,, I)$ is right exact.

3') For every monomorphism $\mu:I\to B$, there exists a morphism $\delta:B\to I$ such that $\delta\mu=1_I$.

4') $I$ is a direct factor (or summand) in every module which contains $I$ as a submodule.

I know how to show that $1\Rightarrow 2\Rightarrow 3\Rightarrow 4$ and $1'\Rightarrow 2'\Rightarrow 3'\Rightarrow 4'$. How would I show that $4\Rightarrow 1$ and $4'\Rightarrow 1'$? Or prove the equivalence in any other way.

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  • $\begingroup$ What is the definition of an injective module in this book? $\endgroup$
    – Bernard
    Feb 18, 2015 at 23:45
  • $\begingroup$ @Bernard the universal property one. $\endgroup$
    – hjhjhj57
    Feb 19, 2015 at 0:21
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    $\begingroup$ Be careful with the term "universal property". A projective object is not universal. $\endgroup$
    – MooS
    Feb 19, 2015 at 6:44
  • $\begingroup$ @MooS Thanks for the clarification. The definition I'm using is that $P$ is projective if given an epimorphism $\psi:M\to N$ and any morphism $f:P\to N$, there exists a morphism (not necessarily unique), $g:P\to M$, such that $f = \psi\circ g$. $\endgroup$
    – hjhjhj57
    Feb 19, 2015 at 7:10

1 Answer 1

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For $4 \Rightarrow 1$ assume $P$ satisfies $4$ and let $M \twoheadrightarrow N$ be surjective with a map $P \to N$. The pullback $M \times_N P \to P$ is surjective (you can prove this categorically or from the definition of the object using that $M \twoheadrightarrow N$ is surjective. Alternatively, see An Introduction to Homological Algebra by Rotman, Exercise 5.10 on page 227) so by $4$ it is just a projection to a summand isomorphic to $P$, say $M\times_N P \cong P\oplus Q$. Then the inclusion of that summand, and the other half of the pullback square ($P \to P\oplus Q \cong M \times_N P \to M \to N$) gives that $P \to N$ factors through $M$. Thus $P$ is projective (I assume you're using the definition stating that $P$ is projective if given an epimorphism $ψ:M→N$ and any morphism $f:P→N$, there exists a morphism (not necessarily unique) $g:P→M$ such that $f=ψ\circ g$).

I suspect a dual argument using the pushforward works for $4' \Rightarrow 1'$, but I haven't thought about it.

Edit: Note that $i_P\colon P \to M \times_N P$ splits $\pi_2$, so $\pi_2\circ i_P = \mathrm{id}_P$. Then $f = f\circ\pi_2\circ i_P = \phi\circ\pi_1\circ i_P$ so $f$ factors through $\phi$.

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  • $\begingroup$ Just out of curiosity, how did you come up with the idea of using pullbacks and pushouts? On another note, proving the surjectivity of the upper arrow in the pullback is quite laborious if I remember correctly, but I believe you can find it in Awodey's book. $\endgroup$
    – hjhjhj57
    Feb 19, 2015 at 2:01
  • $\begingroup$ PS. Not sure about Awodey's anymore. An electronic reference would be great. $\endgroup$
    – hjhjhj57
    Feb 19, 2015 at 4:35
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    $\begingroup$ @hjhjhj57: Was just thinking of what maps come for free and surject onto $P$. As for proving that the arrow is surjective, if you work from the definition of $M \times_N P$ as a module (instead of the categorical definition) then it's trivial to show that the map to $P$ is surjective. $\endgroup$
    – Jim
    Feb 19, 2015 at 5:54
  • $\begingroup$ Great! Could it be that it's as simple as this: Consider $p\in P$, then there exists an $n\in N$ such that $p\mapsto n$. Finally, using that the other map is surjective we get the result? $\endgroup$
    – hjhjhj57
    Feb 19, 2015 at 6:03
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    $\begingroup$ Yep, that's all there is to it. $\endgroup$
    – Jim
    Feb 19, 2015 at 6:17

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