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This is what I got so far before getting stuck

Proof: Show that $im \iota \subseteq ker f$. Let $g \in im \ \iota$. Then there is some $n \in ker\ f$ such that $g = \iota(n)$. If $f: M \rightarrow N$ is surjective, then $im \ \iota = M$. Since $g \in im \ \iota$, then $f(g) = f(\iota(n)) = 0$ implies $g \in ker f$. So $im \ \iota \subseteq ker f$. Next we show that $ker f \subseteq im \ \iota$. Let $h \in ker f$, then the definition of $ker f$ means that $h \in im \ \iota$ if $f$ is surjective.

In the problem I used $\iota $ as the map from $0 \rightarrow ker f$. I also used the definitions of both the $im \iota$ and $ker f$. My question is whether if this outline makes any sense or if there's an appropriate way to approach the problem. I'm not sure if I should approach the relations $imf = ker \iota'$, where $\iota': N \rightarrow co ker f$. Similarly but using $co ker f$ instead of $ker f$. Confusion arises when I use the definition of exactness, i.e. $im \ \iota = ker \ f$ and then there's the $ker \ f $ in the exact sequence $0 \rightarrow ker f \rightarrow M \rightarrow N$. Are these two $ker f$ the same? Sorry this feels like a dumb question.

Alexander King
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1 Answers1

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The inclusion map $\iota:\ker(f)\rightarrow M$ and the quotient map $\mathcal{Q}:N\rightarrow \text{coker}(f)$ should do the trick.

First look at $0\rightarrow\ker(f)\overset{\iota}{\rightarrow} M$. The first arrow is the zero map, whose image is $\{0\}$ and this is the same as the kernel of the inclusion map $\iota$.

Now look at $\ker(f)\overset{\iota}{\rightarrow}M\overset{f}{\rightarrow}N$. The image of $\iota$ is precisely $\ker(f)$, so there is nothing to check here. (To answer your question, the $\ker(f)$ in the exact sequence is the kernel of the given map $f:M\rightarrow N$).

Have a go at the last part, just by computing the image and kernel of the relevant maps.

user505379
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