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Let M: $C^\infty \rightarrow C^\infty$ be the map M:$f \rightarrow tf$ i.e.multiplication by the function t.Show that M is injective, not surjective.

It's very obvious that this map is injective, but how do we prove it's not surjective? Please help me, thank you in advance

Okay, I think that t needs to be non-zero functions. Thus, the only element in the kernel is 0 therefore M is injective. To prove M is not surjective, we need to consider a constant function, $f(x)=c$, which does not have a preimage because $c\over t$ is not in $C^\infty$.

Thank you guys

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What is a feature of every function in the range of $M$? – hardmath Feb 21 '13 at 1:24
@TrevorWilson: Right, it is a bit more work. – hardmath Feb 21 '13 at 1:31
What exactly is $t$? Any function? A constant function? An injective non-zero everywhere function? – Asaf Karagila Feb 21 '13 at 1:32
@user63219: You can make $t$ be a lot of things that make sense, the only question is which one of those is the one you actually need. It would seem most people interpreted it to be the identity function; more explicitly, $M(f(t)) = tf(t)$. – Javier Feb 21 '13 at 2:24
If $t$ is a nowhere $0$ smooth function, then, I guess, $M$ is going to be surjective. – Berci Feb 21 '13 at 11:29

While $0$ is in the domain of these functions, the constant $1$ function is not going to be present in the range of $M$ (because $1/t$ is not defined in $0$).

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Ah, is $t$ a constant? Oh, sorry, I thought, the variable.. – Berci Feb 21 '13 at 1:31
I also became uncertain:) – Berci Feb 21 '13 at 1:33
I assume the mapping $M$ is on $C^\infty (\mathbb{R})$ and that function $t$ is the identity function on $\mathbb{R}$. But the question is certainly open to interpretation. – hardmath Feb 21 '13 at 1:34

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