Let $X, Y, Z$ be arbitrary sets. Suppose $\alpha$ is a function from $X$ to $Y$ and $\beta$ is a function from $Y$ to $Z$ such that $\beta\circ\alpha$ is an onto function. How do I prove that $\beta$ is an onto function? I always get confused when it's proving.
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For any $z\in Z$ there exist $x\in X$ such that $(\beta\circ\alpha)(x)=z$, since $\beta\circ\alpha$ is onto. Now let $y:=\alpha(x)$ we have $\beta(y)=z$ with $y\in Y$, hence $\beta $ is onto.
Math137
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Can you explain me how did you come to the conclusion that "we have β(y)=z with y∈Y," for y:=α(x) – Techwatch Jan 20 '15 at 05:10
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Since $\beta(y)=\beta(\alpha(x))=z$, and we have $x\in X$ so $\alpha(x)\in\alpha(X)\subset Y$ – Math137 Jan 20 '15 at 10:23