[proofplan]
We prove the equality by proving both inclusions inside $\mathfrak g=T_{e_G}G$. If $X$ is tangent to the kernel subgroup, then the one-parameter subgroup generated by $X$ lies in $K$, so composing it with $\varphi$ gives the constant curve at $e_H$ and hence $d\varphi_{e_G}(X)=0$. Conversely, if $d\varphi_{e_G}(X)=0$, exponential naturality gives $\varphi(\exp_G(tX))=\exp_H(0)=e_H$ for every real $t$, so the generated one-parameter subgroup lies in $K$; the closed-subgroup characterization of the [Lie algebra](/page/Lie%20Algebra) then puts $X$ in $\operatorname{Lie}(K)$.
[/proofplan]
[step:Show that tangent vectors to the kernel are killed by the differential]
Let $K:=\ker\varphi$, and let $\iota:K\hookrightarrow G$ denote the inclusion map. Let $X_K\in\operatorname{Lie}(K)=T_{e_K}K$, and write $X:=d\iota_{e_K}(X_K)\in\mathfrak g$ for the corresponding vector under the identification in the statement.
The exponential maps are the smooth maps $\exp_K:T_{e_K}K\to K$ and $\exp_G:\mathfrak g\to G$. Define
\begin{align*}
\beta:\mathbb R&\to K
\end{align*}
\begin{align*}
t&\mapsto \exp_K(tX_K).
\end{align*}
By [citetheorem:8802], applied to the inclusion homomorphism $\iota:K\to G$ and the vector $X_K\in T_{e_K}K$, one has
\begin{align*}
\iota(\exp_K(tX_K))=\exp_G(t\,d\iota_{e_K}(X_K))=\exp_G(tX)
\end{align*}
for every $t\in\mathbb R$. Thus the map
\begin{align*}
\alpha:\mathbb R&\to G
\end{align*}
\begin{align*}
t&\mapsto \exp_G(tX)
\end{align*}
has image contained in $\iota(K)$, which we regard as $K\subset G$. Since $K=\ker\varphi$, we have $\varphi(\alpha(t))=e_H$ for every $t\in\mathbb R$. Therefore $\varphi\circ\alpha:\mathbb R\to H$ is the constant smooth curve at $e_H$.
Differentiating at $0$ gives
\begin{align*}
d\varphi_{e_G}(X)=\frac{d}{dt}\bigg|_{t=0}\varphi(\exp_G(tX))=0\in T_{e_H}H.
\end{align*}
Hence $\operatorname{Lie}(K)\subseteq\ker(d\varphi_{e_G})$.
[/step]
[step:Show that vectors killed by the differential generate curves in the kernel]
Let $X\in\ker(d\varphi_{e_G})$, so $X\in\mathfrak g$ and $d\varphi_{e_G}(X)=0\in\mathfrak h:=T_{e_H}H$. The exponential maps in this step are $\exp_G:\mathfrak g\to G$ and $\exp_H:\mathfrak h\to H$. Define the smooth curve
\begin{align*}
\gamma_X:\mathbb R&\to G
\end{align*}
\begin{align*}
t&\mapsto \exp_G(tX).
\end{align*}
By [citetheorem:8802], applied to the smooth Lie [group homomorphism](/page/Group%20Homomorphism) $\varphi:G\to H$ and the vector $X\in\mathfrak g$, for every $t\in\mathbb R$ one has
\begin{align*}
\varphi(\exp_G(tX))=\exp_H(t\,d\varphi_{e_G}(X)).
\end{align*}
Since $d\varphi_{e_G}(X)=0$, this becomes
\begin{align*}
\varphi(\exp_G(tX))=\exp_H(0)=e_H.
\end{align*}
Thus $\exp_G(tX)\in\ker\varphi=K$ for every $t\in\mathbb R$.
[guided]
We now prove the reverse inclusion carefully. Start with a vector $X\in\ker(d\varphi_{e_G})$. This means that $X$ is a tangent vector at the identity of $G$ and that the [linear map](/page/Linear%20Map)
\begin{align*}
d\varphi_{e_G}:T_{e_G}G\to T_{e_H}H
\end{align*}
sends $X$ to $0\in T_{e_H}H$.
The curve generated by $X$ is
\begin{align*}
\gamma_X:\mathbb R&\to G
\end{align*}
\begin{align*}
t&\mapsto \exp_G(tX).
\end{align*}
To prove that $X$ lies in the Lie algebra of the subgroup $K=\ker\varphi$, we must show that this one-parameter subgroup stays inside $K$. In concrete terms, we need to prove
\begin{align*}
\varphi(\gamma_X(t))=e_H
\end{align*}
for every $t\in\mathbb R$.
This is exactly what exponential naturality gives. By [citetheorem:8802], because $\varphi:G\to H$ is a Lie group homomorphism and $X\in\mathfrak g$, we have
\begin{align*}
\varphi(\exp_G(tX))=\exp_H(t\,d\varphi_{e_G}(X))
\end{align*}
for every $t\in\mathbb R$. The hypothesis $X\in\ker(d\varphi_{e_G})$ now enters: it says $d\varphi_{e_G}(X)=0$. Therefore
\begin{align*}
\varphi(\exp_G(tX))=\exp_H(0).
\end{align*}
The exponential of the zero tangent vector is the identity element, so
\begin{align*}
\varphi(\exp_G(tX))=e_H.
\end{align*}
Hence $\exp_G(tX)\in K$ for all $t\in\mathbb R$.
[/guided]
[/step]
[step:Use the closed-subgroup characterization to identify the Lie algebra]
We now verify the subgroup characterization needed here. Because $K$ is an embedded Lie subgroup of $G$ and $\iota:K\hookrightarrow G$ is the inclusion, a vector $Y\in\mathfrak g$ lies in $\operatorname{Lie}(K)$, viewed through $d\iota_{e_K}$, if and only if the smooth curve
\begin{align*}
t\mapsto \exp_G(tY)
\end{align*}
has image in $K$ for every $t\in\mathbb R$.
Indeed, if $Y=d\iota_{e_K}(Y_K)$ for some $Y_K\in T_{e_K}K$, then [citetheorem:8802] applied to $\iota$ gives $\exp_G(tY)=\iota(\exp_K(tY_K))\in K$ for every $t\in\mathbb R$. Conversely, if $\exp_G(tY)\in K$ for every $t\in\mathbb R$, then the curve $t\mapsto\exp_G(tY)$ is a smooth curve in the embedded submanifold $K$ through $e_K$, and its derivative at $0$ is $Y$; hence $Y\in T_{e_K}K=\operatorname{Lie}(K)$ under the inclusion identification. Therefore
\begin{align*}
\operatorname{Lie}(K)=\{Y\in\mathfrak g:\exp_G(tY)\in K\text{ for every }t\in\mathbb R\}.
\end{align*}
The preceding step proves that the chosen $X\in\ker(d\varphi_{e_G})$ satisfies this condition, so $X\in\operatorname{Lie}(K)$. Hence
\begin{align*}
\ker(d\varphi_{e_G})\subseteq\operatorname{Lie}(K).
\end{align*}
Combining this inclusion with the first inclusion gives
\begin{align*}
\ker(d\varphi_{e_G})=\operatorname{Lie}(K).
\end{align*}
This proves the theorem.
[/step]
