Consider the following fragment from Folland's book "A course in abstract harmonic analysis":
6.1 The Inducing Construction
Let $G$ be a locally compact group, $H$ a closed subgroup, $q : G \to G / H$ the canonical quotient map, and $\sigma$ a unitary representation of $H$ on $\mathcal{H}_\sigma$. We denote the norm and inner product on $\mathcal{H}_\sigma$ by $\lVert u \rVert_\sigma$ and $\langle u,v \rangle_\sigma$, and we denote by $C \left ( G, \mathcal{H}_\sigma \right )$ the space of continuous functiosn from $G$ to $\mathcal{H}_\sigma$. If $f \in C \left ( G, \mathcal{H}_\sigma \right )$, we shall frequently wish to apply the operators $\sigma(\xi)$ to the values $f(x)$, and to avoid clutter we shall usually write $\sigma(\xi)f(x)$ instead of the more precise $\sigma(\xi)[f(x)]$.
The main ingredient in the inducing construction is the following space of vector-valued functions:
$$\mathcal{F}_0 = \left \{ f \in C \left ( G, \mathcal{H}_\sigma \right ) \ : \ \text{$q \left ( \text{supp } f \right )$ is compact and $f(x \xi) = \sigma(\xi^{-1})f(x)$ for $x \in G$, $\xi \in H$} \right \}.$$
Here is how to produce functiosn in $\mathcal{F}_0$:
6.1 Proposition. If $\alpha : G \to \mathcal{H}_\sigma$ is continuous with compact support, then the function
$$f_\alpha(x) = \int_H \sigma(\eta) \alpha(x \eta) \ d \eta$$
belongs to $\mathcal{F}_0$ and is left uniformly continuous on $G$. Moreover, every element of $\mathcal{F}_0$ is of the form $f_\alpha$ for some $\alpha \in C_c \left ( G, \mathcal{H}_\sigma \right )$.
Proof:
Clearly $\boxed{q(\text{supp } f_\alpha) \subset q(\text{supp } \alpha)}$, and ...
How can we conclude that $q(\operatorname{supp}(f_\alpha))$ is compact from the boxed inclusion? I know that $q(\operatorname{supp}(\alpha))$ is compact, so it suffices to show that the former set is closed. But I don't see why this is true.
Any help/comments/insight is highly appreciated!
It is indeed false that the image of closed subset of $G$ under the quotient map $q$ is closed in $G/H$.
However, here we are dealing with a special closed set. Let $A = \operatorname{supp}(f_\alpha)$. By (left) invariance of the (left) Haar measure, we see that $f_\alpha(x)= 0 \iff f_\alpha(xh) = 0$ for all $h \in H$. In particular, we thus have that $AH\subseteq A$.
Now, invoke the following general fact:
Lemma: If $q: G \to G/H$ is the quotient map and if $A \subseteq G$ is a closed subset such that $AH \subseteq A$, then $q(A)$ is closed in $G/H$.
Proof: Since $q$ is an open map, it suffices to show that $q(A^c) = q(A)^c$. The inclusion $q(A^c)\subseteq q(A)^c$ follows from the condition $AH\subseteq A$ (exercise!) while the other inclusion follows from surjectivity of $q$. The lemma follows. $\quad \square$