Let be a ring with 1.

- If and are short exact sequences of (left unital) modules and and are projective, show that .
- If and are short exact sequences of (left unital) modules and and are injective, show that .

(Side note: two modules and are called *projectively equivalent* if there exist projective modules and such that . Likewise, and are *injectively equivalent* if there are injective modules and such that .)

- Let denote the pullback of and . (See this previous exercise about pullbacks.) We claim that the pullback morphisms and are surjective. To see this, suppose and consider . Since is surjective, there exists such that . Thus , and so . Thus is surjective; a similar argument shows that is surjective. Thus and are short exact sequences. We now claim that there exist module homomorphisms and such that the following diagram of modules commutes.
To this end, we prove a lemma.

Lemma 1: Let and be injective module homomorphisms. If , then there exists a unique module homomorphism such that . Moreover, is injective. Proof: Let . Since , there exists such that . Since is injective, this is unique. Define . Certainly is well defined. Moreover, if and and , then we have , so that . So is a module homomorphism. Moreover, , so that . To see uniqueness, if is a module homomorphism such that , then if , we have , so that . Thus . Finally, if , then , so that . Thus is injective.

Now we claim that is injective. To see this, suppose . Then . Moreover, since , we have . Thus , and so is injective. Moreover, suppose with . Then , so that . Thus we have . By Lemma 1, there exists a unique module homomorphism such that . By a similar argument, exists such that . Moreover, and are injective.

We claim now that and are surjective. To see this, let . Now , so that . Now , and in fact . Clearly , so that is surjective. A similar argument shows that is surjective. In fact, we have and . Because and are projective, the sequences centered at are split. Thus we have .

- Let denote the pushout of and . (See this previous exercise about pushouts.) We claim that the pushout morphisms and are injective. To see this, suppose . Then for some . Since is injective, we have , so that . Thus is injective. Similarly, is injective. This yields the two short exact sequences and . We claim that there exist module homomorphisms and such that the following diagram of modules commutes.
To this end, we prove a lemma.

Lemma 2: If and are surjective module homomorphisms and , then there exists a unique module homomorphism such that . Proof: Define as follows: given , where ( exists since is surjective). Note that if , then , so that . Thus is well defined. Moreover, if and , and and , then . Thus , so that is a module homomorphism. That is clear. To see uniqueness, suppose is a homomorphism such that . Given , there exists such that . Now , so that . Finally, to see that is surjective, note that if then since is surjective there exists with . Let ; then .

Now we claim that is surjective. To see this, let . (Recall that , where .) Note that . So is surjective; likewise is surjective. By Lemma 2, there exist unique surjective module homomorphisms and making the diagram commute.

Finally, we claim that and are injective. To see this, let . Since is surjective, there exists an element such that . Now , so that . Thus . That is, for some . Now in , so that for some . In particular, , so that . Thus , and so is injective. Similarly, is injective. So and . Since and are injective, the middle row and column split, and we have .

## Comments

Why you take X/ker(sigma1)? with sigma1 is inyective, so that has no sence, while ker(sigma1)=0, you are saying that the secuence 0 -> Q -> X -> X -> 0 is exact, wich is not true.

Thanks

You’re right- I think it should be . These proofs are dual to each other, so I wrote the first and then edited it to work for the second. Apparently I forgot to dualize .

I’ll mark this as incomplete until I have a chance to properly fix it.

Thanks!

Okay- it should be fixed now. Let me know if there are other errors.

Thanks!