Show there is an exact sequence of $R$-module $\ker(gf) \to \ker(g) \to \mathrm{coker}(f)$

Question

Show there is an exact sequence of $R$-modules $\ker(gf) \to \ker(g) \to \mathrm{coker}(f)$ where $f:M \to N$ and $g:N \to P$ are $R$-module homomorphisms.

I want to avoid using the snake lemma (since I cannot assume it) because this is essentially proving a smaller version of it.

So I need to show $\ker b=$ im $a$ where $a: \ker(gf) \to \ker(g)$ and $b: \ker(g) \to \mathrm{coker}(f)$.

Now $\ker(b) = \{ x\in \ker(g) \mid \mathrm{coker}(f)=0\}$ i.e. when $N=$ im(f) im(a)$=\{y \ \in \ker(g) \mid \exists x \in \ker(gf)\ s.t. y=a(x)\}$

Any hint how I can show these are equal?

Answer 1

By definition, $\operatorname{coker}(f)=N/f(M)$, so you can consider the restriction of the canonical projection $N\to N/f(M)$ to $\ker(g)$, call it $\beta$.

Consider also the map $\alpha\colon\ker(gf)\to\ker(g)$, where $\alpha(x)=f(x)$, that is, the restriction of $f$ to $\ker(gf)$. This is well defined as $x\in\ker(gf)$ implies $f(x)\in\ker(g)$.

You want to prove that the image of $\alpha$ equals the kernel of $\beta$.

By definition, $\beta\alpha=0$, which proves one inclusion.

Suppose $y\in\ker(\beta)$. This means, by definition, $y=f(x)$, for some $x\in M$. Since $y\in\ker(g)$, we have $0=g(y)=gf(x)$, so $x\in\ker(gf)$.

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Some step by step guidance.

The map $\beta\colon\ker g\to N/f(M)=\operatorname{coker}(f)$ is defined by $\beta(y)=y+f(M)$ (the restriction to $\ker(g)\subseteq N$ of the projection map $N\to N/f(M)$).

The map $\alpha\colon \ker(gf)\to \ker(g)$ is defined by $\alpha(x)=f(x)$. This is well defined, because if $x\in\ker(gf)$, then $gf(x)=0$, so $f(x)\in\ker(g)$. It is obviously a homomorphism.

Suppose $y\in\operatorname{im}(\alpha)$. This means $y=f(x)$, for some $x\in\ker(gf)$. Therefore $\beta(y)=y+f(M)=f(x)+f(M)=0+f(M)$, so $y\in\ker(\beta)$.

Conversely, suppose $y\in\ker(\beta)$. This means $y+f(M)=0+f(M)$, that is, $y\in f(M)$. Hence $y=f(x)$, for some $x\in M$. Since $y\in\ker(g)$ by assumption, we also have $g(y)=0$, hence $gf(x)=0$. Therefore $x\in\ker(gf)$ and so $y=\alpha(x)$ (by definition of $\alpha$). Consequently $y\in\operatorname{im}(\alpha)$.