If $T$ is a continuous linear transformation of a normed linear space $N$ into a normed linear space $N'$, and if $M$ is its null space, show that $T$ induces a natural linear transformation $T'$ of $N/M$ into $N'$ and that $\|T'\|=\|T\|$.
This is a question from the book of G. Simmons with the title "Introduction to Topology and Modern Analysis". My guess for $T'$ is $T'(x+M)=T(x)$, which is clearly a linear transformation and well-defined since $M$ is the null space of $T$. I have also managed to see that $\|T'\|\geq \|T\|$. However, I could not see the other direction of the inequality.
You should probably define a norm on $N/M$ as $||x+M||_{N/M} := \inf_{m\in M}||x + m||_N$ which is a norm and well defined because M is closed.
Your guess is correct: $T'(x+M) = T(x)$ is indeed well defined and linear.
To prove that it's bounded take $x+M \in N/M$:
$$ \begin{align} ||T'(x+M)||_{N'} &= ||T(x)||_{N'} \\ &= \inf_{m\in M} ||T(x)||_{N'} \\ &\le \inf_{m\in M} ||T||\, ||x||_N \\ &= ||T|| \inf_{m\in M} ||x+m||_{N} \\ &= ||T||\, ||x+M||_{N/M} \end{align} $$
so $T'$ is bounded and $ ||T'|| \le ||T|| $.
Similarly, take $x\in N$: $$ \begin{align} ||T(x)||_{N'} &= ||T'(x+M)||_{N'} \\ &\le ||T'||\, ||x+M||_{N/M} \\ &= ||T'||\inf_{m\in M} ||x + m||_N \\ &\le ||T'||\, ||x||\,\, \text{(as 0 }\in M) \end{align} $$
which proves that $ ||T|| \le ||T'|| $ and so $ ||T'|| = ||T|| $