# Pull back of $0$-tensor field

Let $$F:M\to N$$ be a smooth map between smooth manifolds. Let $$f$$ be real value continuous function on $$N$$.

Prove that the pull back map has the relationship: $$F^*f = f\circ F$$

Where the pull back map between covariant k-tensor field for each point $$p$$ on manifold is defined as $$(F^*A)_p(v_1,...,v_k) = A_{F(k)}(dF_p(v_1),...,dF_p(v_k)).$$

Since real continuous function is $$0$$-tensor field then $$k = 0$$ is empty so I don't know how to deal with this case?

Ignoring the input vectors $$v_1,...,v_k$$, your definition of the pull-back becomes:
$$(F^* A)_p = A_{F(p)}.$$
If you identify 0-forms with functions (i.e. $$\omega_p = \omega(p)$$), you get
$$(F^* A)(p) = A(F(p)) \; \forall p \in M \implies F^*A = A \circ F.$$
You just wrote it in the definition. When you write $$A_{F(k)}$$ s.t. $$p=F(k)$$. Since you have no vectors as entries you are done.