The following is taken from "Linear Algebra and Geometry" by Leung.
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$\textbf{Exercuse}$ Show that for each subspace $X'$ of a linear space $X,$ the sequence
$$0\xrightarrow{}X'\xrightarrow{i}X\xrightarrow{\pi}X/X'\xrightarrow{}0$$
where $i$ is the inclusion mapping and $\pi$ the natural surjection, is exact.
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For the above exercise, let's relabel the above sequence as: $0\xrightarrow{f_1}X'\xrightarrow{i}X\xrightarrow{\pi}X/X'\xrightarrow{f_2}0.$ We know that $\text{Im }f_1=\text{ker }i=f_i(0)=0,$ and so $\text{Im} f_1=X'.$ We also have $\text{Im } \pi=\pi(X)=X/X'=\text{ker }f_2.$ Since $i$ is injective and being a linear transformation means that $\text{ker } i=0$ and $\text{Im } i=i(X')=X.$ But how do I show that $\text{Im } i=\text{ker }\pi?$
Thank you in advance