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