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Suppose $M,N$ are finite $R$-modules and $a\in R$ is $M$-regular. We then have a short exact sequence $$ 0 \to M \xrightarrow{a} M \to M/aM \to 0$$ which induces the long exact sequence $$ \dots \to \text{Ext}_R^i(M/aM,N) \to \text{Ext}_R^i(M,N) \xrightarrow{a} \text{Ext}_R^i(M,N) \to \dots$$

The question I have is why the map remains multiplication by $a$. The case when $i=0$ is clear, but for larger dimensions I am confused. What is confusing me is that the map on Ext is induced by the maps on the projective resolutions, which as far as I can tell have no relation to the original maps. What am I missing?

As explained in my comment below, taking a projective resolution $P$ of $M$, if $P\to P$ is multiplication by $a$ then we see that $\text{Ext}(a,N) = a$. However, how do we know this is the map that fits into the exact sequence above? Because I do not think $P\to P$ is injective, and the LES is induced by a SES of projective resolutions of $M$ and $M/aM$.

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    Actually, there is a big hint in http://math.stackexchange.com/questions/496797/operatornameann-ra-operatornameann-rb-subseteq-operatornameann-r-operat – 2017-02-23
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    Probably will be marked as duplicate. But with their idea in mind we just take a projective resolution $P$ of $M$ and see that $P \xrightarrow{a} P$ makes the diagram commute. – 2017-02-23
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    That is the obvious way to proceed, and an easy one, to boot. – 2017-02-23
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    @MarianoSuárez-Álvarez but will this be the map used in the LES of Ext? Wouldn't it need to be injective on each level? – 2017-02-23
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    Regarding your question in the last paragraph: you should review how a morphism $f:M\to M'$ of modules induces a map $f^*:Ext(M',N)\to Ext(M,N)$. – 2017-02-23
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    @MarianoSuárez-Álvarez I think I see it now. The map is unique up to homotopy, and homotopic maps yield the same map on homology. Thanks. – 2017-02-23
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    But that is irrelevant, really. You have a map, multiplication by $a$, and then you lift it to the resolutions. You notice that you can lift it by the map which is just muliplication by $a$. The induced map on homology is the one that comes from this. Homotopy invariance means you can ue another lifting, but that is pointless here! – 2017-02-23

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