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Gael Meigniez
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If "strict fibre" just means the fibre over $*$, then youYou can make the construction as follows. There are twothree steps.

First make a Serre fibration over the unit interval, pi:T->I=[0,1] such that the fibre over 0 (resp. 1) is F (resp. F'). You will find how to make this in the general case; for example, for F=1 point and F'=the interval, here is a proof that the triangle does it.

Consider in the plane the triangle T: 0<=y<=x<=1 and the projection pi(x,y)=x. Given a CW-complex A, a continuous map from A to T:a->(x_0(a),y_0(a)) and a homotopy (x_t(a)) (0<=t<=1), consider K={(a,t)\in AxI/x_t(a)=0} and K_0=(Ax0)\cap K. On the complement (Ax0)\K_0 you have the slope function s_0(a):=y_0(a)/x_0(a); extend it to a continuous function s from (AxI)\K to I; the wanted homotopy is (a,t)->(x_t(a),s(a,t)x_t(a)) for (a,t)\notin K and (a,t)->(0,0) for (a,t)\in K. Hence, pi is a Serre fibration.

Second, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$. Then

Third, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; consider the mapping cylinder $C $ of a homotopy equivalence $F\rightarrow F'$ and the projection $pr:C\rightarrow [0,1]$ with $pr^{-1}(1)=F'$; define define $E'$ over $N$ as the amalgamated product of $C$$T$ with $N$ over $G$$pi$ and $g$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want.

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because $C$ is not a Serre fibration over $I=[0,1]$, in general. Instead of $C$, you should use a Serre fibration over $I$ whose fibres over $0$ and $1$ are $F$ and $F'$ respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

If "strict fibre" just means the fibre over $*$, then you can make the construction as follows. There are two steps.

First, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$. Then, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; consider the mapping cylinder $C $ of a homotopy equivalence $F\rightarrow F'$ and the projection $pr:C\rightarrow [0,1]$ with $pr^{-1}(1)=F'$; define $E'$ over $N$ as the amalgamated product of $C$ with $N$ over $G$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want.

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because $C$ is not a Serre fibration over $I=[0,1]$, in general. Instead of $C$, you should use a Serre fibration over $I$ whose fibres over $0$ and $1$ are $F$ and $F'$ respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

You can make the construction as follows. There are three steps.

First make a Serre fibration over the unit interval, pi:T->I=[0,1] such that the fibre over 0 (resp. 1) is F (resp. F'). You will find how to make this in the general case; for example, for F=1 point and F'=the interval, here is a proof that the triangle does it.

Consider in the plane the triangle T: 0<=y<=x<=1 and the projection pi(x,y)=x. Given a CW-complex A, a continuous map from A to T:a->(x_0(a),y_0(a)) and a homotopy (x_t(a)) (0<=t<=1), consider K={(a,t)\in AxI/x_t(a)=0} and K_0=(Ax0)\cap K. On the complement (Ax0)\K_0 you have the slope function s_0(a):=y_0(a)/x_0(a); extend it to a continuous function s from (AxI)\K to I; the wanted homotopy is (a,t)->(x_t(a),s(a,t)x_t(a)) for (a,t)\notin K and (a,t)->(0,0) for (a,t)\in K. Hence, pi is a Serre fibration.

Second, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$.

Third, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; define $E'$ over $N$ as the amalgamated product of $T$ with $N$ over $pi$ and $g$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want.

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes)

If "strict fibre" just means the fibre over *$*$, then you can make the construction as follows. There are two steps. First

First, let f:X->X$f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood N$N$ of * onto *$*$ onto $*$. Pulling back your fibration p$p$ through f$f$, you are reduced to the case where p $p$ is a projection FxN->N$F\times N\rightarrow N$ over N$N$. Then, consider a function g: N->I$g: N\rightarrow I$ whose value is 0$0$ on partial N$\partial N$ and 1$1$ on *; [consider$*$; consider the mapping cylinder C$C $ of a homotopy equivalence F->F' equivalence $F\rightarrow F'$ and the projection pr:C->[0,1]$pr:C\rightarrow [0,1]$ with pr^-1(1)=F'];$pr^{-1}(1)=F'$; define E'$E'$ over N$N$ as the amalgamated product of C$C$ with N$N$ over G;$G$; define E'$E'$ as E$E$ over (X-N)$(X-N)$. ifIf you have chosen N$N$ to retract by deformation on *$*$, then this E'$E'$ will have the property you want. Maybe

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because C$C$ is not a Serre fibration over I=[0,1]$I=[0,1]$, in general. Instead of C$C$, you should use a Serre fibration over I$I$ whose fibres over 0$0$ and 1$1$ are F$F$ and F'$F'$ respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

If "strict fibre" just means the fibre over *, then you can make the construction as follows. There are two steps. First, let f:X->X be homotopic to the identity and contract some neighborhood N of * onto *. Pulling back your fibration p through f, you are reduced to the case where p is a projection FxN->N over N. Then, consider a function g: N->I whose value is 0 on partial N and 1 on *; [consider the mapping cylinder C of a homotopy equivalence F->F' and the projection pr:C->[0,1] with pr^-1(1)=F']; define E' over N as the amalgamated product of C with N over G; define E' as E over (X-N). if you have chosen N to retract by deformation on *, then this E' will have the property you want. Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because C is not a Serre fibration over I=[0,1], in general. Instead of C, you should use a Serre fibration over I whose fibres over 0 and 1 are F and F' respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

If "strict fibre" just means the fibre over $*$, then you can make the construction as follows. There are two steps.

First, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$. Then, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; consider the mapping cylinder $C $ of a homotopy equivalence $F\rightarrow F'$ and the projection $pr:C\rightarrow [0,1]$ with $pr^{-1}(1)=F'$; define $E'$ over $N$ as the amalgamated product of $C$ with $N$ over $G$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want.

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because $C$ is not a Serre fibration over $I=[0,1]$, in general. Instead of $C$, you should use a Serre fibration over $I$ whose fibres over $0$ and $1$ are $F$ and $F'$ respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

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Gael Meigniez
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If "strict fibre" just means the fibre over *, then you can make the construction as follows. There are two steps. First, let f:X->X be homotopic to the identity and contract some neighborhood N of * onto *. Pulling back your fibration p through f, you are reduced to the case where p is a projection FxN->N over N. Then, consider a function g: N->I whose value is 0 on partial N and 1 on *; consider[consider the mapping cylinder C of a homotopy equivalence F->F' and the projection pr:C->[0,1] with pr^-1(1)=F';=F']; define E' over N as the amalgamated product of C with N over G; define E' as E over (X-N). if you have chosen N to retract by deformation on *, then this E' will have the property you want. Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because C is not a Serre fibration over I=[0,1], in general. Instead of C, you should use a Serre fibration over I whose fibres over 0 and 1 are F and F' respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

If "strict fibre" just means the fibre over *, then you can make the construction as follows. There are two steps. First, let f:X->X be homotopic to the identity and contract some neighborhood N of * onto *. Pulling back your fibration p through f, you are reduced to the case where p is a projection FxN->N over N. Then, consider a function g: N->I whose value is 0 on partial N and 1 on *; consider the mapping cylinder C of a homotopy equivalence F->F' and the projection pr:C->[0,1] with pr^-1(1)=F'; define E' over N as the amalgamated product of C with N over G; define E' as E over (X-N). if you have chosen N to retract by deformation on *, then this E' will have the property you want. Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes)

If "strict fibre" just means the fibre over *, then you can make the construction as follows. There are two steps. First, let f:X->X be homotopic to the identity and contract some neighborhood N of * onto *. Pulling back your fibration p through f, you are reduced to the case where p is a projection FxN->N over N. Then, consider a function g: N->I whose value is 0 on partial N and 1 on *; [consider the mapping cylinder C of a homotopy equivalence F->F' and the projection pr:C->[0,1] with pr^-1(1)=F']; define E' over N as the amalgamated product of C with N over G; define E' as E over (X-N). if you have chosen N to retract by deformation on *, then this E' will have the property you want. Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes) Sorry, I come back to this answer which I realize is not quite correct, because C is not a Serre fibration over I=[0,1], in general. Instead of C, you should use a Serre fibration over I whose fibres over 0 and 1 are F and F' respectively; there should be a canonical construction of this kind, but right now I cannot be more precise. Hope it helps.

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Gael Meigniez
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