Something is puzzling me here...
Suppose $U$, $V$ and $W$ are finite dimensional vector spaces of different dimensions. Let $\phi_1:U \rightarrow V$, $\phi_2:V \rightarrow W$ and $\phi_3:U \rightarrow W$ be linear maps.
Since $U$, $V$ and $W$ are of different dimensions, $\phi_1$, $\phi_2$ and $\phi_3$ are not invertible (in the sense that for each $i$, there exists no linear map $\phi^{-1}_i$ such that $\phi^{-1}_i$ is both a left and right inverse to $\phi_i$).
Now suppose $$\phi_2 \phi_1 = \phi_3$$ and that the map $\phi_2$ is injective. A linear map is injective if and only if it has a left inverse. Thus, there exists a map $\phi^{-1}_2$ such that $\phi^{-1}_2\phi_2=1$. So $$\phi^{-1}_2\phi_2 \phi_1 = \phi^{-1}_2\phi_3$$ and hence $$\phi_1 = \phi^{-1}_2\phi_3.$$ If we multiply on the left again by $\phi_2$ then we have $$\phi_2\phi_1 = \phi_2\phi^{-1}_2\phi_3=\phi_3.$$ Does this not imply that $\phi_2\phi^{-1}_2$ is an identity map? But then surely this implies that $\phi_2$ has a right inverse, which is true if and only if $\phi_2$ is surjective. This would imply that $\phi_2$ is bijective, which is false, since $\mathrm{dim}(V)\neq\mathrm{dim}(W)$.
Where is the error in my reasoning?