Proposition 3.2.7 of Bland on short exact sequences, part 2

Question

I am reading the book of Bland, Rings and their Modules (2011). My question is related to proposition 3.2.7 on page 85.

Given a short exact sequence $S_1: 0\longrightarrow M_1\stackrel{\alpha}\longrightarrow M\stackrel{\beta}\longrightarrow M_2\longrightarrow 0$ of left or right $R$-modules and given $M\cong M_1\oplus M_2$. $R$ may commutative or noncommutative.

I am looking for an example in which $\operatorname{im} \alpha = \ker \beta$ is NOT a direct summand of $M$. (Thus $S_1$ cannot be split.)

This question is related to these two posts:

Proposition 3.2.7 of Bland on short exact sequences

Example of a non-splitting exact sequence $0 → M → M\oplus N → N → 0$

It seems like the second post you've cited answers your question, no? — Mr. Chip, Aug 11 '16 at 10:37
Possible duplicate of Example of a non-splitting exact sequence $0 → M → M\oplus N → N → 0$ — Pierre-Guy Plamondon, Aug 11 '16 at 11:01
Maybe, but can you prove that im f is not a direct summand of M? @Mr Chip, I don't see the relation with my problem, I am talking about modules and not about groups, abelian or not. — Steenis, Aug 11 '16 at 11:06
First, abelian groups are $\mathbb{Z}$-modules. Second, the link on the splitting lemma is valid for any abelian category; in particular, it works for any module category. Third, the splitting lemma will tell you that the image of $\alpha$ is a direct summand of $M$ if and only if the sequence is split. Hence, if you have already shown that a sequence is not split, then you have your example. — Pierre-Guy Plamondon, Aug 11 '16 at 11:11
But! In the Splitting lemma of Mr. Chip is this valid: S1 is left split iff Im G is direct summand of M? In prop 3.2.7 of Bland (1), (2) and (3) are not equivalent. In my view, this contradicts the Splitting Lemma. — Steenis, Aug 11 '16 at 11:21
Of course, you are right, Pierre-Guy, But I am very curious for a direct proof. I think it will enlighten my understanding. Maybe it makes no sense, but I want to know. — Steenis, Aug 11 '16 at 11:24
It is not a contradiction if it is given that q and r are the natural inclusion and projection, respectively. — Steenis, Aug 11 '16 at 12:36

Proposition 3.2.7 of Bland on short exact sequences, part 2

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