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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?

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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

$$ (F^* A)(p) = A(F(p)) \; \forall p \in M \implies F^*A = A \circ F. $$

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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.

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