How to prove that $A^T y=0\implies y=0$ if and only if $A$ defines a surjective linear transformation?

Question

I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $\text{coker} f=\{0\}$ if and only if $f$ is surjective?

I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).

Thank you

Answer 1

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m \times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

Answer 2

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

Answer 3

So, we have a function $f: X\rightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $\rightarrow$ $A$ has a left inverse $\rightarrow$ $A^T$ has a right inverse $\rightarrow$ $A$ is surjective.