Is the Lie algebra morphism induced by injective/surjective Lie group morphism also injective/surjective? I want to show that if $f$ is an injective/surjective Lie group homomorphism than $f_*$ is injective/surjective.
$\require{AMScd}$
$$\begin{CD}G     @>f>>   H\\
@AexpAA         @AAexpA\\
Lie(G)     @>>f_*>   Lie(H)
\end{CD}$$
Take open sets $U,V,U', V'$ such that $V=exp(U), V'=exp(U'), f(V)=V', f_*(U)=U'$ and $exp|_U, exp|_V$ are diffeomorphisms.
injectivity :
Suppose $f_*(x)=f_*(y)$ (where $f_*(x),f_*(y)\in V$. Then take $t$ small enough such that $tx,ty\in U$. Then by the diagram, we have $exp^{-1}(f(exp(tx)))=exp^{-1}(f(exp(ty)))$ so $tx=ty$ i.e. $x=y$.
surjectivity : Consider $y\in V\subset Lie(H)$. Now chase back in the diagram a preimage of $y$ with respect to $f_*$. Then since every $\hat{y}$ can be written as $ty$ for some $t\in \mathbb{R}$ and some $y \in V$ we are done.
Is this correct
What about the converse? f_* injective/surjective $\implies$ $f$ injective/surjective?
 A: To answer one of your questions: If $f_*$ is surjective and $H$ is connected then $f$ is surjective:
By commutativity of your diagram, $f$ must be surjective onto exp(Lie$(H)$), so the image of $f$ contains an open neighbourhood $U\subseteq H$ of $e\in H$.  This will generate a subgroup $H'\subset H$, which is open:  for any $y\in H'$, we have $yU$ a neighbourhood of $y$ contained in $H'$.
Then each coset of $H'$ is open (as multiplication by $y\in H$ is a homeomorphism), so the union of all non-trivial cosets is open, implying that $H'$ is closed.
We conclude that $H'$ is a non-empty open, closed, subset of a connected space $H$, so $H=H'$ and $H'\subseteq$im$(f)$.
As exemplified in the comments, the requirement that $H$ is connected is necessary. Otherwise, the inclusion of the connected component of $e\in H$ will not be surjective.
To answer another of your questions, even if $G,H$ are both simply connected, $f_*$ injective does not imply that $f$ is injective.  Consider $f\colon \mathbb{R}\to SU(2)$ mapping: $$x\in \mathbb{R}\mapsto \left(\begin{array}{cc}e^{ix}&0\\0&e^{-ix}\end{array}\right)$$
