I don't understand something about the following result:
Claim: The topology of the group $F(X)$ is the finest topological group topology on $F_a(X)$ that generates on $X$ its original topology.
($X$ is a Tychonoff topological space, and $F(X)$ denotes the free topological group on $X$, and $F_a(X)$ denotes the free group on $X$).
Proof: We already know that $F(X)$ is algebraically isomorphic with $F_a(X)$ and the topology $\tau$ of $F(X)$ generates on X its original topology
Now, let $\tau '$ be a group topology on $F_a(X)$ such that $\tau '\big|_X=\tau _X$, where $\tau _X$ is the topology of $X$. Extend the identity mapping $i:X\to X$ to a continuous homomorphism $\overline{i}:F(X)\to F_a(X)$, where $F_a(X)$ is endowed with the group topology $\tau '$. Clearly, $\overline{i}$ is an algebraic isomorphism, and the continuity of $\overline{i}$ implies that $\tau '\subseteq \tau$.
My question is, why does the continuity of $\overline{i}$ imply that $\tau '\subseteq \tau$?
$\bar i$ continuous $\overset{def}\iff$ for all open $U\subseteq F_a(X)$, i.e., $U\in\tau'$ we have $\bar i^{-1}(U)$ is open in $F(X)$, so, as $i$ and by the group structure, $\bar i$ is also identity, $U=\bar i^{-1}(U)$, it means $U\in\tau$.