Span,kernel, rank and dimensions.

Question

I am trying to understand the approach used to solve the following multiple choice problem:

"We consider three vector spaces all defined on a field $K$: $U, V,W$. Let $g: U \rightarrow V, f: V \rightarrow W$ be linear maps and $u_1,...,u_m$ be a basis of $U$. Then, choose 1 of the following:

1. If $dim(U) = dim(W)$ and $ {f◦g(u_1),...,f◦g(u_m)}$ spans $W$, then $rank(g) = m$

2. If $dim(W) \leq dim(U)$, then ${f◦g(u_1),...,f◦g(u_m)}$ spans $W$.

3. If $Ker(f) = \{0\}$, then ${f◦g(u_1),...,f◦g(u_m)}$ is a linearly independent set.

4. If $dim(U) = dim(W)$ and $ {f◦g(u_1),...,f◦g(u_m)}$ spans $W$, then $Ker(f) = \{0\}$"

Now the correct answer is 1, but I have no idea why this is the case. I kind of know why 3 is incorrect (I think its because we are missing the hypothesis on the kernel of g; if it also contains just the 0 vector, then it should be true). Besides that, I am a bit stuck and would appreciate your help.

Answer 1

1. If $f\circ g(u_1),\dots,f\circ g(u_m)$ span $W$, and $\dim W=\dim U=m$, it implies $f\circ g(u_1),\dots,f\circ g(u_m)$ is a basis of $W$, hence they're linearly independent.
2. Not necessarily. For instance, $f$ or $g$ might be the $0$ map.
3. $\ker f=0$ only says $f$ is injective. But if $g(u_1),\dots, g(u_m)$ are not linearly independent, $f\circ g(u_1),\dots,f\circ g(u_m)$ can't be.
4. It's true if $\dim V=\dim U$: same argument as in 1: $f\circ g(u_1),\dots,f\circ g(u_m)$ are linearly independent, hence $g(u_1),\dots, g(u_m)$ are linearly independent. But if $\dim V >\dim U$, it does not imply $f$ is injective.

Counter example: take $V=U\oplus U$, and $g$ be the canonical injection into the first factor, defined by $g(u)=(u,0)$, and $f$ be the projection on the first factor. The $f\circ g=\operatorname{id}_U$, the simplest of isomorphisms, yet $\ker f=\{0\}\oplus U$.