Why is the oldform map injective? Consider the space of cusp forms $S_k(\Gamma_0(N))$; it has two different maps to $S_k(\Gamma_0(Np))$ where $(p, N) = 1$. We can combine them into a map $$S_k(\Gamma_0(N)) \oplus S_k (\Gamma_0(N)) \to S_k(\Gamma_0(Np))$$ given explicitly by $$(f_1, f_2) \mapsto f_1 + f_2 |_k \begin{pmatrix} p & 0 \newline 0 & 1 \end{pmatrix}.$$ I think I've seen it asserted that this map is injective; why?
Here's what I've tried so far. If we define $\pi = \begin{pmatrix} p & 0 \newline 0 & 1\end{pmatrix}$, then an element of the kernel would have to be of the form $(f|_k \pi, -f).$ So somehow if $f|_k \pi$ is still $\Gamma_0(N)$-invariant, it should have to be trivial. 
From an adelic perspective, $f\in S_k(\Gamma_0(N))$ is a function on $GL_2(\mathbb{A}_f)$ that satisfies some transformation properties; in particular, it's constant on the compact open $U$ corresponding to $\Gamma_0(N)$. 
If $f|_k \pi $ is still constant on $U$, then $f$ is constant on both $\pi U \pi^{-1}$ and $U$. I believe that I could conclude if those two compact opens (plus images of $GL_2(\mathbb{Q})$) together generated all of $GL_2(\mathbb{A}_f)$. Do they? Or is there a different reason that $f$ must be trivial?
 A: I don't know much about the adelic perspective, but I am familiar with an argument using the Atkin-Lehner operator and $q$-expansions. This argument is adapted from Proposition 5.1 of https://arxiv.org/abs/1808.04588.
Let $a,b\in\mathbb{Z}$ be such that $pb-aN=1$. The Atkin-Lehner operator $w_p$ on $S_k(\Gamma_0(Np))$ is defined by 
$$w_p(f)=f|_k\begin{pmatrix} p & a \newline Np & pb \end{pmatrix}.$$
Note that $w_p$ is an involution, i.e. $w_p^2(f)=f$. Moreover, if $f\in S_k(\Gamma_0(N))\subseteq S_k(\Gamma_0(Np))$, then $w_p(f)=f|_k\pi$. Let me know if you want me to elaborate on either of these points.
Now as you've stated, if $(f_1,f_2)$ is in the kernel of the oldform map, then $f_1=-f_2|_k\pi=-w_p(f_2)$. So, in this case
$$f_2=w_p^2(f_2)=-w_p(f_1)=-f_1|_k\pi$$
and thus
\begin{align} f_1=-w_p(f_2)=-f_2|_k\pi=f_1|_k\pi^2.\tag{1}\end{align}
If $f_1=\sum_{n=1}^\infty a_nq^n$ then (1) becomes
\begin{equation}\sum_{n=1}^\infty a_nq^n=p^k\sum_{n=1}^\infty a_nq^{np^2}.\tag{2}\end{equation}
It suffices to show that $f_1=0$. So, suppose for a contradiction that $f_1$ is nonzero. Let $n>0$ be the least integer with $a_n\neq 0$. Then the left-hand side of (2) has a $q^n$ term whereas the right-hand side of (2) does not. This is impossible so $f_1$ must be zero as required. 
