# Schanuel's Lemma

I want to prove the following:

Assume that we have two exact sequences of $$R$$-modules $$0 \to K \xrightarrow{\psi} P \xrightarrow{\varphi} M \to 0$$ and $$0 \to K' \xrightarrow{\psi'} P' \xrightarrow{\varphi'} M \to 0$$ where $$P,P'$$ are projective, and we want to prove that $$P \oplus K' \cong P' \oplus K.$$

My reasoning: We have the fiber-product $$X := \{(p,p') \;|\; p \in P, p' \in P', \ \text{such that} \ \varphi(p) = \varphi'(p')\}$$ of $$\varphi$$ and $$\varphi'$$.

We look at the sequence $$0 \to \operatorname{ker}(\pi) \xrightarrow{\iota} X \xrightarrow{\pi} P \to 0.$$

where (I believe) $$\iota:\operatorname{ker}(\pi) \to X$$ is just the injection of elements in $$X$$ that get sent to $$0$$ under $$\pi$$, and $$\pi:X \to P$$ is the surjection from the fiber-product onto $$P$$. Now, since $$\pi$$ is the projection, it is surjective, and clearly the injection $$\iota$$ is injective, hence we get an exact sequence.

Now, since $$P$$ is a projective $$R$$-module, we have that that every short exact sequence splits, so that the there exists a section $$h:P \to X$$ so that $$h(P) \cong P$$ and we have $$X \cong \operatorname{ker}(\pi) \oplus h(P) \cong \operatorname{ker}(\pi) \oplus P.$$

Now, we note that $$\pi:X \twoheadrightarrow P$$ must be defined explicitly (I think) by $$(p,p') \longmapsto p$$ so that $$\operatorname{ker}(\pi) = \{(0,p')\;|\; p' \in P' \ \text{such that} \ \varphi(0) = \varphi'(p')\}$$ hence if $$(0,p') \in \operatorname{ker}(\pi) \implies \varphi'(p') = 0.$$

Now, we take note of the other SES we had defined above $$0 \to K' \to P' \xrightarrow{\varphi'} M \to 0$$ and since it is exact, we have that $$\operatorname{Im}(\psi') = \operatorname{ker}(\varphi')$$ so that for each $$p' \in P'$$ such that $$\varphi'(p') = 0$$ there exists an $$k' \in K'$$ so that $$\psi'(k') = p'$$. And since $$\psi'$$ is injective, we have that if $$\psi'(k'_1) = \psi'(k'_2) \implies k_1' = k_2'$$.

Hence we have (I believe) an isomorphism $$K' \cong \operatorname{ker}(\pi)$$ due to the fact that we earlier concluded that $$\operatorname{ker}(\pi)$$ is exactly elements $$p' \in P'$$ such that $$\varphi'(p') = 0$$.

We see that it follows that $$X \cong \operatorname{ker}(\pi) \oplus P \cong K' \oplus P \underbrace{\cong}_{\text{canonical isomorphism}} P \oplus K'.$$

Now, I assume the second case is done similarly.

Is this correct? If not, what is wrong?

• $$X \cong P’ \oplus K$$ @MarianoSuárez-Álvarez Commented Oct 7, 2023 at 14:12
• .@MarianoSuárez-Álvarez do you agree? The idea is to show that $$X \cong P' \oplus K$$ and $$X \cong P \oplus K'$$ so that $$X \cong P' \oplus K \cong P \oplus K'.$$ Commented Oct 7, 2023 at 19:41

Your claims are, indeed, all true. They follow from more general properties of exact sequences and fibre products.

First, whenever you have surjection $$g\colon B\twoheadrightarrow C$$, you can complete this to an short exact sequence $$0 \to A\xrightarrow{f} B\xrightarrow{g} C\to 0$$ by defining $$A=\ker(g)$$ and letting $$f$$ be the canonical inclusion $$\ker(g)\hookrightarrow B$$.

Second, if you form the fibre product $$X$$ of a surjection $$P\to M$$ with any map $$P'\to M$$, the resulting map $$X\to P'$$ is always surjective. Indeed, if we define $$x$$ explicitly to be $$X=\{(p,p')\in P\times P'\,|\,\varphi(p)=\varphi'(p')\}$$ then $$X\to P$$ and $$X\to P'$$ are just the restrictions of the coordinate projections. In particular, if $$p'\in P'$$ we can let $$m=\varphi'(p')$$ and choose a $$\varphi$$-preimage $$p\in P$$. Then $$(p,p')\in X$$ since $$\varphi(p)=m=\varphi'(p')$$ by construction. As along $$X\to P'$$, $$(p,p')$$ is mapped to $$p'$$, this shows surjectivity.

Third, since we just saw how the map $$\pi\colon X\to P$$ is explicitly constructed, its kernel is naturally isomorphic to $$K'$$. The argument you gave here is perfectly fine.

The second and third point actually generalise beyond the explicit realisation of the fibre product: they follow from the definition of the fibre product as a universal object. If you do not know too much about this kind of stuff, do not worry. It is not important for this particular problem.

One can however avoid (the explicit usage) of fibre products.

By the lifting property of projective moduels, we can complete the given diagram as: $$\require{AMScd}$$ $$\begin{CD} 0 @>>> K @>{\psi}>> P @>{\varphi}>> M @>>> 0 \\ {} @V{\beta}VV @V{\alpha}VV @V{1_M}VV \\ 0 @>>> K'@>>{\psi'}> P' @>>{\varphi'}> M @>>> 0 \end{CD}$$ This uses only projectivity of $$P$$. Now I claim that there is a short exact sequence $$0 \to K \to K'\oplus P \to P' \to 0$$ where we define $$K\to K'\oplus P$$ and $$K'\oplus P\to P'$$ naturally from the abundance of given maps (to not spoil the fun, I will this part to you; let me know if you need any help). Now using only the projectivity of $$P'$$, we deduce that the above sequence is split-exact and therefore $$K'\oplus P\cong K\oplus P'\,.$$ I said above that we avoid the explicit usage of fibre products; indeed, a commuting square as above together with the (appropriate) short exact sequence below is actually equivalent to be a fibre product (even, the fibre product you considered).