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[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r = k^\ast : \Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction along $k:N_0 \to N$.

On the other hand, consider, as in the question, a manifold $V$ obtained from $N$ by gluing an open collar along the boundary of $N$. Then we also have the map defined in the question $$ i = l_\ast : \Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion $l:N \to V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s = k^\ast : \Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j = m_\ast : \Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the left) with the inclusion $m:\interior N \to V$.

Finally, observe that both $s$ and $j$ are homotopy equivalences (one needs to use the collars to show this). Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Hence, the homotopy fibre of $r$ is equivalent to its fibre. The desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r = k^\ast : \Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction along $k:N_0 \to N$.

On the other hand, consider, as in the question, a manifold $V$ obtained from $N$ by gluing an open collar along the boundary of $N$. Then we also have the map defined in the question $$ i = l_\ast : \Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion $l:N \to V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s = k^\ast : \Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j = m_\ast : \Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the left) with the inclusion $m:\interior N \to V$.

Finally, observe that both $s$ and $j$ are homotopy equivalences (one needs to use the collars to show this). Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Hence, the homotopy fibre of $r$ is equivalent to its fibre. The desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r:\Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction. There is also the map from the question $$ i:\Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion of $N$ in $V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s:\Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j:\Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the right) with the inclusion $\interior N \to V$. Note that both $s$ and $j$ are homotopy equivalences. Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Therefore, the homotopy fibre of $r$ is equivalent to its fibre. Finally, the desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r = k^\ast : \Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction along $k:N_0 \to N$.

On the other hand, consider, as in the question, a manifold $V$ obtained from $N$ by gluing an open collar along the boundary of $N$. Then we also have the map defined in the question $$ i = l_\ast : \Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion $l:N \to V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s = k^\ast : \Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j = m_\ast : \Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the left) with the inclusion $m:\interior N \to V$.

Finally, observe that both $s$ and $j$ are homotopy equivalences (one needs to use the collars to show this). Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Hence, the homotopy fibre of $r$ is equivalent to its fibre. The desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map which Tom uses $$ r:\Diff(N) \To \Emb(N_0,\interior N) $$ given by restriction. There is also the map from the question $$ i:\Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion of $N$ in $V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s:\Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j:\Emb(N_0,\interior N) \to \Emb(N_0,V)$ is geiven by composition (on the right) with the inclusion $\interior N \to V$. Note that both $s$ and $j$ are homotopy equivalences. Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Therefore, the homotopy fibre of $r$ is equivalent to its fibre. Finally, the desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r:\Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction. There is also the map from the question $$ i:\Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion of $N$ in $V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s:\Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j:\Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the right) with the inclusion $\interior N \to V$. Note that both $s$ and $j$ are homotopy equivalences. Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Therefore, the homotopy fibre of $r$ is equivalent to its fibre. Finally, the desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.