Tangent space of the image of an immersion

Question

I'm going to phrase my question in a certain context (while I do believe generalisations are possible!). Say we have an injectieve Lie group homomorphism $F\colon G\to H$ between Lie groups $G$ and $H$. Then $\operatorname{Im}F$ is an immersed submanifold of $H$. Is it then true that $$ T_e(\operatorname{Im}F)\subset dF_e(T_e G), $$ where and $T_e(\operatorname{Im}F)$ and $T_e G$ denote the tangent space at the identity for $\operatorname{Im}F$ and $G$ respectively?

EDIT:

In addition to the answer provided below, I wanted to make some remarks (just for myself):

When we write $T_{F(x)}(\operatorname{Im}(F))$, we mean in fact $$ d\iota_{F(x)}T_{F(x)}(\operatorname{Im}(F)), $$ where $\iota\colon\operatorname{Im}(F)\to N$ is the inclusion map. Hence, if we have such an $F\circ\gamma$ (as mentioned in the answer), we can restrict it to $\operatorname{Im}(F)$ and retain smoothness, since $\operatorname{Im}(F)$ is an immersed submanifold. Let's denote this restricted map by $\tilde F\circ\gamma$. The differential $d\iota_{F(x)}$ will then send $(\tilde F\circ\gamma)'(0)$ to $(F\circ\gamma)'(0)$, since $$ \iota\circ\tilde F\circ\gamma=F\circ\gamma, $$ and hence indeed $dF_x(T_x M)\subset T_{F(x)}(\operatorname{Im}F)$.

Answer 1

This is a special case of a more general result:

Theorem Let $f\colon M\to N$ be an immersion and $X=f(M)\subseteq N$ be the image of $f$. If $m\in M$ and $x=f(m)\in X$, then

$$ T_xX = \mathrm{d}f_m (T_mM)$$

This is 'obvious' – immersions are local embeddings meaning that we can take open $U$ around $x$ such that $f|_U$ is an embedding. And for an embedded submanifold this should clearly be the case. But let's make a formal proof.

Proof: Consider a path $\gamma$ representing $u\in T_mM$. Then $\mathrm{d}f_m(u)$ is represented by path $f\circ \gamma$, that is in $X$ from the definition of $X$. Hence we have $\mathrm{d}f_m(T_mM)\subseteq T_xX$. Dimensions of both spaces are equal, what finishes the proof.