It is easy to miss the point that in the second definition $\mathbb Q/\mathbb Z$ is required to be *discrete* in Hiro's question. Hence even if a continuous morphism $f:G\to T$ has image in $\mathbb Q/\mathbb Z$ it cannot automatically be considered as a continuous map $f_0:G\to \mathbb Q/\mathbb Z$ and so might not be a character in the second sense. An example for this failure is to take $G=T_{tors} =\mathbb Q/\mathbb Z\subset T$ , the torsion subgroup of $T$ with its induced topology from the circle and for the character $f $ (in the first sense) the inclusion $f:G\hookrightarrow T$. Even though $G$ is torsion, the corestricted morphism $f_0:G\to \mathbb Q/\mathbb Z$ is not a character in the second sense, since it is not continuous. However if $G$ is compact, its image under a character is discrete in the circle, hence finite, and the two concepts coincide. This applies in particular to profinite groups (since they are automatically compact).