A function $f: R \rightarrow S$ is simply a unique “mapping” of elements in the set $R$ to elements in the set $S$. (b) Prove that if TS is surjective, then T is surjective. If $$f$$ is a linear map between vector spaces (and not just an arbitrary function between sets), there is a simple way to check if $$f$$ is injective. Deﬁnition 5. 2 If dim(V) V and T: V -> W be two linear maps. This is not onto because this guy, he's a member of the co-domain, but he's not a member of the image or the range. Injective, Surjective and Bijective "Injective, Surjective and Bijective" tells us about how a function behaves. The diﬀerentiation map T : P(F) → P(F) is surjective since rangeT = P(F). Proof. In particular, ker(T) = f0gif and only if T is bijective. 1 If dim(V) >dim(W), then T is not injective. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share …