Theorem 1.3. Let $\sigma : U\to V$ be a linear map from an $m$-dimensional vector space to an $n$-dimensional one. The rank $\rho$ of $\sigma$ satisfies $\rho(\sigma)\le \min\{m,n\}$.
Prove: If $\sigma : U\to V$ and $\tau : V\to W$ are linear maps, then the rank of the composition satisfies $\rho(\tau\sigma)\leq min\{\rho(\tau),\rho(\sigma)\}$.
Proof:
Define the following linear transformations $\sigma$ and $\tau$ such that $\sigma:U\rightarrow V$ and $\tau: V\rightarrow W$. Let $\rho$ mean rank and let $dimU=m$, $dimV=n$, $dimW=q$. By Theorem 1.3., \begin{align} \rho(\tau\sigma)\leq min\{m,q\}; \rho(\sigma)\leq min\{m,n\}; \rho(\tau)\leq min\{\rho(\sigma),q\} \end{align} Thus, $\rho(\sigma)\leq m$ and $\rho(\tau)\leq q$. Therefore, $\rho(\tau\sigma)\leq min\{\rho(\tau),\rho(\sigma)\}$.
This statement is not true: $\rho(\tau)\le \min\{\rho(\sigma),q\}$.
What is true, is that for the restriction to the image of $\sigma$, $\rho(\tau|_{\sigma(U)})\le \operatorname{dim}(\sigma(U))$. This is what you need to make the proof work.