I am trying to show $q: \text{fimm}(M,N)\to \text{Map}(M,N), (f,\delta f)\mapsto f$ is a Serre fibration. Here $\text{fimm}(M,N)$ consists of pairs $(f,\delta f)$ such that $f: M\to N$ is a continous map and $\delta f: TM\to f^*TN$ is a vector bundle map such that it is injective on each fiber.
I have tried the following. Consider the commutative diagram:
$$\require{AMScd}$$ \begin{CD} X\times\{0\}@>h>> \text{fimm}(M,N)\\ @V{incl}VV @VVqV\\ X\times[0,1]@>>F> \text{Map}(M,N) \end{CD}
We need to construct a lift $H: X\times [0,1]\to \text{fimm}(M,N)$ such that $H_{|X\times \{0\}}= h$ and $q\circ H= F$. A candidate is defined by $$ H(x,t)= (F(x,t), \delta h(x,0) ). $$ By $\delta h(x,0)$ we mean the linear injection obtained in the second argument of $h(x,0)=(f,\delta f)$, but now it sends elements from $T_pM$ to $T_{F(x,t)(p)}N$ instead of $T_{f(p)}N$. It is continuous because both its arguments are. Trivially $q\circ H= F$. Additionally, we have by commutativity of the diagram that the map $F(x,0)$ is the same as the one in the first component of $h(x,0)$ and then $\delta h(x,0)$ also coincides with the linear injection given by $h(x,0)$. Thus we have $H_{|X\times \{0\}}= h$, this means $q$ is a Hurewicz fibration.
Is this correct?