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Let $U,V,W$ be vector spaces and $g:U\rightarrow V,h:V \rightarrow W$ be linear mappings. Prove that the composite linear mapping $h\circ g:U\rightarrow W$ satisfies $h\circ g=0$ if and only if $ \operatorname{im}(g) \subseteq \ker(h)$.

This statement seems obvious but I am terrible at proving things so I have no idea how to even begin.

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  • $\begingroup$ Hint: What dies it mean that $\;(h\circ g)(x)=0$? $\endgroup$ – Bernard May 5 at 10:05
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I will show you one direction: Let $\text{im}(g) \subseteq \text{ker}(h)$. For $ u \in U$, we have $g(u) \in \text{im}(g)$, so $h(g(u)) = 0$ since $\text{im}(g) \subseteq \text{ker}(h)$. Try the other direction, you can do it.

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  • $\begingroup$ Let $h\circ g=0$. This means that for all $v \in \operatorname{im}(g), h(v)=0$ which implies that $\operatorname{im}(g) \subseteq \ker(h)$? Is this correct? $\endgroup$ – Montes May 5 at 10:14
  • $\begingroup$ yap, good job!! $\endgroup$ – Riquelme May 5 at 10:49

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