Even if you're only interested in say cohomology with coefficients in the constant sheaf, working with constructible sheaves gives you extra flexibility and is more amenable to inductive proofs.

Here is a basic theorem in the topology of algebraic varieties one of whose proofs could serve as a motivation. I discussed it in some MO answer, maybe I'll link it later, and I learned this from Lazarsfeld's book "Positivity in Algebraic Geometry" (volume 1).

**Theorem.** Let $X\subseteq \mathbf{C}^n$ be an affine algebraic variety, i.e. a closed subset cut out by polynomial equations. Then $H^i(X, \mathbf{Z}) = 0$ for $i>\dim X$.

Here $\dim X$ is the algebraic dimension, can be defined in various ways, see any textbook on algebraic geometry. The "dimension" (in whatever sense) of $X$ as a topological space is thus $2\dim X$. The above result is really special to affine varieties: the projective space $\mathbf{C}P^n$ has dimension $n$ and nonzero cohomology in even degrees $i\leq 2n$.

One proof of the above result uses Morse theory (a careful analysis of $X$ by means of an auxiliary function $\mathbf{C}^n\to [0,\infty)$ such as $\sum |z_i|^2$), and shows a bit more: $X$ is homotopy equivalent to a CW complex with cells in dimensions $\leq n$ (the Andreotti-Frankel theorem). This is not the proof I'd like to mention.

The proof which generalizes well to other contexts, e.g. to $\ell$-adic cohomology, due to Michael Artin, proceeds by induction on $d=\dim X$. By the Noether normalization lemma, there exists a finite morphism
$$ f\colon X\to \mathbf{C}^d $$
("finite" here is equivalent to "proper with finite fibers"). How is the cohomology of $X$ related to cohomology of $\mathbf{C}^d$? You really need sheaf cohomology to answer that. In this case (because the map is finite!) we obtain isomorphisms
$$ H^i(X, \mathbf{Z}) \simeq H^i(\mathbf{C}^d, f_*\mathbf{Z}). $$
The sheaf $f_*\mathbf{Z}$ on the right hand side is no longer the constant sheaf, but can be shown to be a constructible sheaf. (Over a Zariski dense open subset of $\mathbf{C}^n$, it will be a locally constant sheaf of rank $e$ where $e$ is the degree of the finite map.) Therefore the theorem will follow from a more general statement below.

**Theorem.** Let $F$ be a sheaf on $\mathbf{C}^d$, constructible with respect to a Zariski stratification. Then $H^i(\mathbf{C}^d, F)=0$ for $i>d$.

Now we are able to proceed by induction on $d$: we take a linear projection
$$ p\colon \mathbf{C}^d \to \mathbf{C}^{d-1}. $$
If $p$ is generic, then the cohomology of $F$ fits inside a long exact sequence (a form of the Leray spectral sequence):
$$ \cdots \to H^i(\mathbf{C}^{d-1}, p_* F)\to H^i(\mathbf{C}^d, F) \to H^{i-1}(\mathbf{C}^{d-1}, R^1 p_* F)\to \cdots $$
Here $R^1 p_* F$ is the first higher push-forward, and (again for $p$ generic) the sheaves $p_* F = R^0 p_* f$ and $R^1 p_* F$ will have stalks which compute the cohomology of the fibers:
$$ (R^i p_* F)_y = H^i(p^{-1}(y), F). $$
(I am telling this backwards: one shows the above formula for all $i$, and since we know the theorem for $d=1$, we know that $R^i p_* F = 0$ for $i>1$, and then we get the long exact sequence.) One can show that $p_* F$ and $R^1 p_* F$ are again constructible with respect to an algebraic stratification, and we proceed by induction. (The case $d=1$ still has to be done by hand.)

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