Is limit of null-homotopic maps null-homotopic? The question is motivated by my failed comment to this one.
Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds).
Let $\varphi_{n}:M\to N$ be null-homotopic and convergent to $\varphi:M\to N$ in the compact open topology.

Does it follow that $\varphi$ is null-homotopic?

Note that homotopy between maps is a path in $C(M,N)$ (for nice $M$, $N$), and so what my question asks is whether the path component of a constant map is closed in $C(M,N)$.
I am waaay out of my depth here, but perhaps there is a continuous or positive lower semi-continuous functional on $C(M,N)$ akin to the topological degree such that null-homotopic maps would be the zero-set of that functional?
 A: I'll provide a general theorem (then one has to apply it to specific circumstances). There is a micro-dictionary/Notation at the bottom of this note.
B-assumption:   Space $\ N\times N\times[0;1]\ $ is normal.
Every metric space $\ N\ $ satisfies B-assumption.
Notation   Let $\ \mathcal W_N\ $ be the set of all closed
neighborhoods of diagonal $\ \Delta_N\ :=\ \{(y\ y):\ y\in N\}\ $
in $\ N\times N.$
Family $\ \mathcal W_N\ $ is a basis of all neighborhoods of the diagonal $\ \Delta_N.$
A-assumption:   Space $\ N\ $ is an ANR, meaning that
for every normal space $\ X\ $ and closed subset $\ A\ $ of $\ X,\ $
and for every continuous function $\ f:A\to N\ $ there exists
a neighborhood $\ U\ $ of $\ A\ $ and continuous $\ F:U\to N\ $ such
that $\ F|A=f.$
Thus, $\ N^2\ $ is an ANR too.
Definition:   Sequence $\ f_n:M\to N\ $ is d-convergent to
$\ f:M\to N\ \Leftarrow:\Rightarrow $
$$ \forall_{V\in\mathcal W_N}\exists_{m\in\Bbb N}
     \forall_{n\ge m}\quad (f_n\triangle f)(M)\,
       \subseteq V $$
Only continuous functions are meant:
THEOREM   Let sequence $\ f_n:M\to N\ $ be d-convergent
to $\ f:M\to N.\ $ Then there exists $\ m\in\Bbb N\ $ such that
$\ f_n\ $ and $\ f\ $ are homotopic for every $\ n\ge m.$
PROOF   Diagonal $\ \Delta_N\ $ is an ANR because it is
homeomorphic to $\ N.\ $ Also, $\ \Delta_N\ $ is closed in $\ N^2\ $ since $\ N\ $ is Hausdorff.  Thus, there exists $\ U\in\mathcal W_N\ $
and a retraction $\ \rho:U\to\Delta_N\ $ (it is an extension of the
identity map on $\ \Delta_N.)$
Consider the function $\ g\ $ from a closed subset of $\ N^2\times[0;1]\ $ into $\ N^2\ $ given as follows:

*

*$\ \forall_{y\in N^2}\quad g(y\ 0)\ :=\ y; $

*$\ \forall_{y\in\Delta_N}\forall_{t\in[0;1]}
         \quad g(y\ t)\ := y; $

*$\ \forall_{y\in U}\qquad g(y\ 1)\ :=\ \rho(y). $
The arguments of $\ g\ $ belong to the union of three closed subsets
of $\ N\times[0;1],\ $ where the three parts of the definition of
$\ g\ $ coincide on the overlaps hence $\ g\ $ is well defined. This
$\ g\ $ admits an extension  $\ G_0\ $ over a closed neighborhood of its closed $3$-part domain. This neighborhood includes $\ V\times[0;1],\ $ where $\ V\subseteq U\ $ is a closed neighborhood of $\ \Delta_N,\ $
because $\ [0;1]\ $ is compact.
Now, by (very elementary and great) Borsuk's homotopy extension
theorem, there is homotopy
$$ H:N^2\times[0;1]\to N^2 $$
such that:

*

*$\ \forall_{y\in N^2}\qquad H(y\ 0)\ :=\ y; $

*$\ \forall_{y\in V}\forall_{t\in[0;1]}
         \quad H(y\ t)\ := G_0(y\ t); $
Let $\ m\in\Bbb N\ $ and $\ n\ge m\ $ be as in
Definition. Let homotopies
$\ h_n\ h:M\times[0;1]\to N\ $ be given as
$$ h_n\ :=\ \pi'\circ H\circ
          ((f_n\triangle f)\times\Bbb I );$$
$$ h\ :=\ \pi''\circ H\circ
          ((f_n\triangle f)\times\Bbb I );$$
where $\ \pi'\ \pi'':N^2\to N\ $ are the canonical projections,
and $\ \Bbb I:[0;1]\to[0;1]\ $ is the identity map.
We see that:
$$ \forall_{x\in M}\quad h_n(x\ 0)\ =\ f_n(x); $$
$$ \forall_{x\in M}\quad h(x\ 0)\ =\ f(x); $$
$$ \forall_{x\in M}\quad h_n(x\ 1)\ =\ h(x\ 1). $$
Define $\ \gamma_n:M\to Y\ $ by $\ \gamma_n(x):=h_n(x\ 1)=h(x\ 1).\ $
We see that $\ f_n\ $ is homotopic to $\ \gamma_n\ $ is homotopic
to $\ f.\,\ $ Remember (observe) that $\ H\ $ in the expressions for
$\ h_n(x\ 1)\ $ and $\ h(x\ 1)\ $ is equal to $\ G_0\ $ (we have
$\ (f_n(x)\ f(x))\in V).\ $   End of PROOF

NOTATION

*

*For functions $\ f:P\to Q\ $ and $\ g:R\to S,\ $ the cartesian product $\ f\times g:P\times Q\to R\times S\ $ is given by
$$ \forall_{(p\ r)\in P\times R}\quad
         (f\times g)(p\ r)\ :=\ (f(p)\ g(r)\,) $$

*Let $\ P=R\ $ and $\ \Delta_P:=\{(p\ p): p\in P\}.\ $ Then
$\ f\triangle g: P\to Q\times S\ $ is given as follows:
$$ f\triangle g\ := (f\times g)\circ \delta_P $$
where $\ \delta_P:P\to P\times P\ $, and
$\ \forall_{p\in P}\ \delta_P(p):=(p\ p).$
A: Please see the answer to Annie's question.
Non-density of continuous functions to interior in set of all continuous functions
