The composition of surjective linear transformations is surjective

Created almost 9 years ago, updated about 8 hours ago

If $T:U\to V$ and $S:V\to W$ are surjective, each $w\in W$ equals $S(v)$ for some $v$, and $v=T(u)$ for some $u$. Thus $w=(S\circ T)(u)$. The converse only forces $S$ to be surjective.

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