Topology Generated by Homomorphisms into a Topological Group

Question

Let $G$ be a group, and let $\Gamma$ be a topological group. Give $G$ the coarsest (initial?) topology with respect to which all homomorphisms from $G$ to $\Gamma$ are continuous. My question is,

Does such a topology turn $G$ into a topological group? If not, what if there were an embedding (injective homomorphism) from $G$ to $\Gamma$? What if $\Gamma$ were a Lie group?

EDIT

Not sure how relevant this is, but here is something to consider. Note that $\Bbb{Q}$ and $\Bbb{R}$ are topological groups with their standard topologies. A simple group theory exercise shows that all homomorphisms from $\Bbb{Q}$ to $\Bbb{R}$ are of the form $q \mapsto mq$ for some $m \in \Bbb{R}$. All of these homomorphisms are continuous, but none of them are open maps (right?).

I don't think this resolves the question in any way, but it suddenly occurred to me, so I included it.

Answer 1

Yes, it makes $G$ a topological group. Indeed, let $X$ be the set of all homomorphisms $G\to\Gamma$. Endow $\Gamma^X$ with the product topology. Let $u:G\to\Gamma^X$ be the canonical homomorphism (namely $u(g)=(f(g))_{f\in X}$. Let $\tau$ be the pull-back by $u$ of the topology of $\Gamma^X$. Then $\tau$ is the desired topology on $G$.

Indeed each $f$ is continuous from $(G,\tau)$ to $\Gamma$ (since $u$ is continuous and $f=u\circ \pi_f$ where $\pi_f$ is the projection on the $f$-component $\Gamma^X\to\Gamma$), and by definition the continuity of each $f$ forces the continuity of $u$, which means containing $\tau$.

Notes:

1. the closure $H$ of $u(G)$ in $\Gamma^X$ also provides a natural "$\Gamma$-completion" of $G$.
2. this works in a broader context, replacing $\Gamma$ with a family $\mathcal{G}$ of topological groups. For instance, when $\mathcal{G}$ is the family of all finite discrete groups, $\tau$ is known the profinite topology on $G$, and the above $H$ (previous note) is known as the profinite completion of $G$.