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Suppose that we have tree spaces $A, B, C$ (let say CW-complexes). Are given two pairs of maps:

$$f_{0},g_{0}: A\rightarrow B $$ $$f_{1}, g_{1}: A\rightarrow C $$ such that $f_{0},g_{0}$ are homotopic and $f_{1},g_{1}$ are also homotopic.

Is it true that $colim [B\leftarrow^{f_{0}} A\rightarrow^{f_{1}} C] $ and $colim[B\leftarrow^{g_{0}} A\rightarrow^{g_{1}} C]$ are weakly equivalent ?

Moreover, we assume that $g_{0}= i\circ f_{0}$ and $g_{1}=j\circ f_{1}$ where $i:B\rightarrow B $ and $j: C\rightarrow C$ are weak equivalences.

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    $\begingroup$ This is true for homotopy colimits. That's more or less why the notion of homotopy colimit was invented, to fix this defect of the usual colimit. $\endgroup$ Commented Jun 30, 2018 at 21:44

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No. Take $A=[0,1]$, $B=S^1$, $C=\text{point}$. There is only one possible choice for $f_1$ and $g_1$. There are many possible choices for $f_0$ and $g_0$, but they are all homotopic. Choose $f_0$ to be constant, and choose $g_0$ to be surjective. Then the pushout for $f$ is $S^1$, and the pushout for $g$ is a point.

UPDATE: Here is an adjusted answer for the adjusted question. Use the same spaces as before, but with the model $[0,1]/(0\sim 1)$ for $B=S^1$. Define $f_0\colon A\to B$ by $f_0(t)=[t/2]$, and define $g_0$ by $g_0(t)=[t]$. Define $h_s\colon B\to B$ by $h_s([t])=[\min(1,(1+s)t)]$, so $h_0=1$ and $h_1\circ f_0=g_0$. Again the pushout for $f$ is $S^1$, and the pushout for $g$ is a point.

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  • $\begingroup$ Sorry I have just seen your answer, I a added a condition in my question. Is there still an obvious counterexample ? Thanks. $\endgroup$
    – mathphys
    Commented Jun 30, 2018 at 20:53
  • $\begingroup$ May be I'm missing something, but in your example $h_{1}$ is not a weak equivalence as it was assumed in the question. right ? $\endgroup$
    – mathphys
    Commented Jul 1, 2018 at 14:26
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    $\begingroup$ $h_1$ is homotopic to the identity, so of course it is a weak equivalence. $\endgroup$ Commented Jul 1, 2018 at 14:37

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