Exactness of $\textrm{Ind}_H^G$ Let $G$ be a group of td type (Hausdorff, and every neighborhood of the identity contains a compact open subgroup).  A smooth representation of $G$ is an abstract representation $(\pi,V)$ such that the stabilizer of any vector is open.
Let $H$ be a closed subgroup of $G$, and $(\sigma,W)$ a representation of $H$.  Define $\textrm{Ind}_H^G(\sigma)$ to be the smooth representation of $G$ consisting of all functions $f: G \rightarrow W$ such that $f(hg) = \sigma(h) f(g)$ for all $h \in H, g \in G$, and such that $f$ is right $K$-invariant for some open compact subgroup $K$ of $G$: $f(gk) = f(g)$.  Here $G$ acts on $\textrm{Ind}_H^G(\sigma)$ by right translation: $g \cdot f(x) = f(xg)$.
To any $H$-linear map of smooth representations $\phi: (W,\sigma) \rightarrow (U, \tau)$, define $$\textrm{Ind}_H^G(\phi): \textrm{Ind}_H^G(\sigma) \rightarrow \textrm{Ind}_H^G(\tau)$$
by $\textrm{Ind}_H^G(\phi)(f)(g) = \phi(f(g))$.  Then $\textrm{Ind}_H^G$ is an additive functor from the category of smooth $H$-representations into that of smooth $G$-representations.
Supposedly $\textrm{Ind}_H^G$ is an exact functor.  Why is this?
 A: Frobenius reciprocity says that $\textrm{Ind}_H^G$ is a right adjoint to the restriction functor $\textrm{Res}_H^G$.  That is, for smooth representations $\pi$ of $G$ and $\sigma$ of $H$, there is a bijection, natural in $\pi$ and $\sigma$:
$$\textrm{Hom}_G(\pi, \textrm{Ind}_H^G(\sigma)) = \textrm{Hom}_H(\textrm{Res}_H^G \pi, \sigma)$$
Taking this for granted, we get that $\textrm{Ind}_H^G$ is left exact as a right adjoint.  The difficult part is then showing that $\textrm{Ind}_H^G$ preserves surjective maps.
The point is this: any smooth representation of a compact group of td type is semisimple, which is to say a direct sum of irreducible representations, or equivalently spanned by irreducible representations.  It can be shown from here that if $(\pi,V)$ is a smooth representation of $G$, and $\rho$ is an irreducible representation of a fixed compact open subgroup $K$ of $G$, then $V$ is the direct sum of $K$-stable spaces $V(\rho)$, where $V(\rho)$ is the $\rho$-isotypic component of $V$: the subspace of $V$ spanned by all $K$-subspaces which are $K$-isomorphic to $\rho$.
Any $G$-map of smooth representations $V \rightarrow V'$ is in fact a direct sum of $K$-maps of $K$-spaces $V(\rho) \rightarrow V'(\rho)$.  In particular, if $v' \in V'$ is a $K$-fixed vector (that is, $v'$ lies in the isotypic space corresponding to the trivial representation) lying in the image of $V$, then $v'$ is mapped to by a $K$-fixed vector of $V$.
So, let $\phi: (W,\sigma) \rightarrow (U,\tau)$ be a surjective map of smooth $H$-representations.  Let $f \in \textrm{Ind}_H^G(\tau)$, and let $K$ be a compact open subgroup of $G$ for which $f$ is right $K$-invariant.
Break $G$ up into orbits under the action of $H \times K$ (by $(h,k) \cdot g = hgk^{-1}$).  Let $I$ be the set of orbits, and choose for each $i \in I$ an orbit representative $g_i$ of $i$.  It is easy to see that $f(g_i) \in U$ is fixed by the compact open subgroup $H \cap g_iKg_i^{-1}$ of $H$.  Since $\phi$ is surjective, and by the "in particular" above, we can choose a vector $w_i \in W$, also fixed by $H \cap g_iKg_i^{-1}$, such that $\phi(w_i) = f(g_i)$.  We then define a function
$$f_0: G \rightarrow W$$
by specifying its values on each of the orbits $Hg_iK$:
$$f_0(hg_ik) = \sigma(h)w_i$$
To show that this is well defined, we need to show that for a fixed orbit $i$, if $h_1g_i k_1 = hg_ik$, then $\sigma(h_1)w_i = \sigma(h)w_i$.  This follows because $\sigma(h^{-1}h_1)$ stabilizes $w_i$, as $h^{-1}h_1 \in H \cap g_iKg_i^{-1}$.
It is immediate that $f_0$ lies in $\textrm{Ind}_H^G(\sigma)$.  Finally, $\textrm{Ind}_H^G(\sigma)(f_0) = f$, since 
$$\textrm{Ind}_H^G(\sigma)(f_0)(hg_ik) = \phi(f_0(hg_ik)) = \phi(\sigma(h)w_i) = \tau(h)f(g_i) = f(hg_i) = f(hg_ik)$$
