Let $S$ be a subspace of $V$ and let $\tau\in\mathcal{L}(V,W)$ satisfy $S\subset\mathrm{ker}(\tau)$. Then there's a unique linear map $\tau':V/S\to W$ s.t. $\tau'\circ\pi_S=\tau$. Moreover, $\mathrm{ker}(\tau')=\mathrm{ker}(\tau)/S$ and $\mathrm{im}(\tau')=\mathrm{im}(\tau)$.
Here $\pi_S$ is such that $\pi_S(v) = v+S$, $S$ being a subspace of $V$. This proof is taken from the textbook Advanced Linear Algebra by S. Roman. I'm gonna display part of it, until the part my question arises. It goes as follows:
We define $\tau'$ by the condition $\tau'\circ\pi_S = \tau$. This function is well defined iff $v+S=u+S \implies \tau'(v+S) = \tau'(u+S)$ which is equivalent to each of the following statements: $$ \begin{align} v+S = u+ S &\implies\tau v = \tau u\\ v-u\in S &\implies \tau(v-u) = 0\\ x\in S &\implies \tau x = 0\\ S&\subset\mathrm{ker}(\tau) \end{align} $$
I'm gonna stop the proof right here, because it is here that my question arises. In the previous statements, why is it $\tau$ instead of $\tau'$? I ask this because $\pi_S v = \pi_S u \implies v-u \in S \iff \tau'(u-v)=0 \iff \tau'(S)=\{0\} \implies S\subset\mathrm{ker}(\tau')$.
The proof is wrong. $v+S=u+S\implies\tau'(v+S)=\tau'(u+S)$ isn’t what needs to be proved; it’s a tautology that a function applied to equal elements yields equal results. What needs to be proved is actually the first of the displayed sequence of statements, because this is what ensures that it makes sense to say “We define $\tau'$ by the condition $\tau'\circ\pi_S = \tau$.”