If $f\circ g$ is onto, does it follow that $f$ is onto?
I know what onto means: for every b, there is an a such that f(a) = b.
I have no idea how to apply that to this question. Pls help!!
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Yes, it follows that $f$ is onto: Assuming $g\colon X \to Y$ and $f\colon Y\to Z$ with $f\circ g$ onto $Z$, it follows that $f$ is onto $Z$. Suppose $z\in Z$. Then there is $x\in X$ with $f(g(x)) = z$, but then $y = g(x)\in Y$ has image $z = f(y)$ under $f$. So $Z = range(f\circ g) \subseteq range(f) \subseteq Z$, so $range(f) = Z$.
To see it another way: the range of $f\circ g$ is the range of $f\!\restriction\! range(g)$ ($f$ restricted to $range(g)\subseteq Y$). This restriction of $f$ is a subset of $f$, therefore its range is a subset of the range of $f$.
$g$ might not be onto $Y$, but it fills up just enough $Y$: every $z\in Z$ comes from some $x\in X$ via $f\circ g$, and then $z$ comes from $y=g(x) \in range(g)\subseteq Y$.
BrianO
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very confusing... – satjav Nov 08 '15 at 10:35
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How's that? what's unclear? – BrianO Nov 08 '15 at 10:36
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Any clearer yet? – BrianO Nov 09 '15 at 04:45
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Yes I get it now, but this answer is wrong. If f is 2x and g is x/2, f ∘ g is x. – satjav Nov 09 '15 at 04:49
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y = 2x is not onto because u can never get an odd number for y. But y =x is onto. – satjav Nov 09 '15 at 04:50
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Are you talking about the integers or about the reals? You're confusing the two. As real functions, of course $f(g(x)) = x$, but then $f(x) = 2x$ is a bijection of the reals, so it's onto. HOWEVER it seems you mean functions from integers to integers ("odd numbers", etc.), and then "x/2" is not what you think: it's floor(x/2), i.e. $\lfloor x/2\rfloor$, so $g(3) = g(2) = 1, g(1)=g(0)=0$, and then $f\circ g$ is NOT the identity, it's not even onto. Really, the answer is correct. – BrianO Nov 09 '15 at 05:01