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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.

hjhjhj57
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  • What is the definition of an injective module in this book? – Bernard Feb 18 '15 at 23:45
  • @Bernard the universal property one. – hjhjhj57 Feb 19 '15 at 00:21
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    Be careful with the term "universal property". A projective object is not universal. – MooS Feb 19 '15 at 06:44
  • @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 $f = \psi\circ g$. – hjhjhj57 Feb 19 '15 at 07:10

1 Answers1

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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 $\phi$.

Pablo S. Ocal
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Jim
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  • 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.

     – hjhjhj57 Feb 19 '15 at 02:01
  • PS. Not sure about Awodey's anymore. An electronic reference would be great. – hjhjhj57 Feb 19 '15 at 04:35
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    @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. – Jim Feb 19 '15 at 05:54
  • 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? – hjhjhj57 Feb 19 '15 at 06:03
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    Yep, that's all there is to it. – Jim Feb 19 '15 at 06:17
  • After a close inspection, I'm not sure the approach works. The problem is that once we have the inclusion, the triangle we get isn't commutative. Let $p\in P$, then $$ p\overset{i_P}{\mapsto}(0,p)\overset{\pi_1}{\mapsto} 0\mapsto 0\neq f(p). $$ – hjhjhj57 Feb 19 '15 at 19:50
  • What is $\pi_1$ and $i_P$? – Jim Feb 19 '15 at 19:54
  • Let me name all the arrows: By hypothesis we have the arrows $\varphi:M\twoheadrightarrow N$ and $f:P\to N$. Since the category of modules has pullbacks, we also have two maps: $\pi_1: M\times_N P\to M$ given by $\pi_1(m,p)=m$ and $\pi_2: M\times_N P\twoheadrightarrow P$ given by $\pi_2(m,p)=p$. In my previous comment, $i_P:P\to M\times_N P\simeq \ker \pi_2\oplus P$ is the canonical inclusion (this isomorphism is a consequence of your answer). – hjhjhj57 Feb 19 '15 at 19:58
  • You are assuming $\ker\pi_2 \subseteq \ker\pi_1$, which is not necessarily true. Remember that the decomposition $\ker\pi_2 \oplus P$ is not the same as having $M \times_N P \subseteq M \times N$. – Jim Feb 19 '15 at 20:05
  • I'm not sure that's the problem. We want to show that $\varphi\circ (\pi_1\circ i_P) =f$, right? So taking $p\in P$, on one side we get that $$(\varphi\circ (\pi_1\circ i_P))(p)=\varphi (\pi_1(0,p)) = \varphi(0) = 0.$$ While $f(p)$ isn't necessarily $0$. – hjhjhj57 Feb 19 '15 at 20:09
  • When you write $i_P = (0, p)$ this $(0, p)$ is an element of $\ker\pi_2 \oplus P$. This is isomorphic to $M \times_N P$ and under that isomorphism $(0, p)$ does not necessarily go to $(0, p)$. Then when you write $\pi_1(0, p) = 0$ you are treating $(0, p)$ as an element of $M \times_N P$. – Jim Feb 19 '15 at 20:13
  • I see your point. But being a morphism wouldn't it map $(0,p)$ to some $(0,p')$? On the other hand, if we use that $\varphi\circ\pi_1 = f\circ\pi_2$ wouldn't we get the desired result? – hjhjhj57 Feb 19 '15 at 20:17
  • It doesn't map $(0, p)$ to $(0, p')$. In fact $(0, p) \in \ker\pi_2 \oplus P$ maps to $(m, p) \in M \times_N P$ where $m$ satisfies $\phi(m) = f(p)$. Note that if $(0, p') \in M \times_N P$ then by definition $f(p') = \phi(0) = 0$. So you are essentially assuming that $i_P$ maps $P$ into $\ker\pi_2 \oplus \ker f$. – Jim Feb 19 '15 at 20:19
  • Great. I'll review the details and your edit. I appreciate the useful discussion. – hjhjhj57 Feb 19 '15 at 20:21