Let $(L,\tau)$ be a topological vector space over $\Bbb{C}$ and $M$ be a subspace of $L$ and let $$f:L\to L/M$$ be the canonical map of $L$ onto $L/M$.Let $ \hat \tau$ be the quotient topology on $L/M$. Thus $U \subset L/M$ is open iff $f^{-1}(U)$ is open in $L$. I would like to show that $\hat \tau$ is translation-invariant. It suffices to show that for any $\hat a\in L/M$, $U+\hat a$ is open in $L/M$ if $U\in\hat\tau$.
Let $\hat a\in L/M$ and $U\subset L/M$ be open. By definition of $\hat \tau$, $ f^{-1}(U)$ is open in $L$. Need to show that $f^{-1}(U+\hat a)$ is also open in $L$. If one has
$$
f^{-1} (U+\hat a)= f^{-1} (U) + f^{-1}(\{\hat a\}),
$$
then the proof will be done since $f^{-1} (U) + f^{-1}(\{\hat a\})$ is open in $L$ by the translation-invariance property of $\tau$. I can only show that
$$
f^{-1} (U+\hat a)\supset f^{-1} (U) + f^{-1}(\{\hat a\})
$$
using the linearity of $f$. How about the other direction? Or could anyone come up with an alternative approach?
Let $f(x) \in U + \hat a$, that is, $x + M \in U + \hat a$, so for some $\hat u = u + M \in U$, $x + M = \hat u + \hat a$. That is, there is $m \in M$, $a \in \hat a$, such that $x = u + a + m$. Now $u \in f^{-1}(U)$, as $f(u) = \hat u$, $a+m \in f^{-1}(\{\hat a\}) = \hat a$, that is $x \in f^{-1}(U) + f^{-1}(\{\hat a\})$.