Let $(X,A)$ be a CW-pair. Let $f_1,f_2 : A \longrightarrow X_0$ be such that $f_1 \simeq f_2.$ Then $$X_0 \cup_{f_1} X \simeq X_0 \cup_{f_2} X.$$
Proof $:$ Let $F : A \times I \longrightarrow X_0$ be a homotopy between $f_1$ and $f_2.$ Let $Z = X_0 \cup_F (X \times I),Z_1 = X_0 \cup_{f_1} X$ and $Z_2 = X_0 \cup_{f_2} X.$ Then $Z$ contains $Z_1$ and $Z_2$ as subspaces. Now since $(X,A)$ is a CW-pair $(X \times \{0\}) \cup (A \times I)$ is a deformation retract of $X \times I.$ Let $H : X \times I \times I \longrightarrow X \times I$ be that deformation retraction. Then it will induce a homotopy $\widetilde H : Z \times I \longrightarrow Z$ between $Z_1$ and $Z$ which is identity on $X_0.$ Analogously we can invert the homotopy to get a homotopy between $Z_2$ and $Z$ which is also identity on $X_0.$ Since homotopy is a transitive relation we find that $Z_1$ is homotopic to $Z_2$ relative to $X_0.$
I don't understand how $\widetilde {H}$ can be induced from $H.$ Also what is the meaning of inverting a homotopy? Would anybody please help me in this regard? Thanks in advance.
EDIT $:$ I think that $\widetilde H$ gives a deformation retraction of $Z$ onto $Z_1$ relative to $X_0.$ But I don't know how to obtain that. At the end I think it is shown that the spaces $Z_1$ and $Z_2$ are homotopy equivalent. But I can't consider it to be well explained proof.
We have following pushout diagram $:$
$$\require{AMScd} \begin{CD} A \times I @>{}>{\text {inclusion}}> X \times I \\ @V{F}VV @VV{}V\\ X_0 @>{}>{\text {inclusion}}> Z \end{CD}$$
Then it will induce the following pushout $:$ $$\require{AMScd} \begin{CD} (A \times I) \times I @>{}>{}> (X \times I) \times I \\ @V{F \times {\text {id}_I}}VV @VV{}V\\ X_0 \times I @>{}>{}> Z \times I \end{CD}$$ So by universal property of pushout, in order to get a map from $Z \times I$ to $Z$ it is enough to get maps $j_1 : (X \times I) \times I \longrightarrow Z$ and $j_2 : X_0 \times I \longrightarrow Z$ such that $$j_1 \big\rvert_{(A \times I) \times I} = j_2 \circ (F \times \text {id}_I).$$
I can guess what would be $j_1.$ I think it will be the composition of the following maps $:$
$$(X \times I) \times I \xrightarrow {H} X \times I \hookrightarrow X_0 \sqcup (X \times I) \xrightarrow {\text {quotient map}} Z.$$ I think $j_2 = q \circ p,$ where $p : X_0 \times I \longrightarrow X_0$ is the projection and $q : X_0 \longrightarrow Z$ is the quotient map. Then we have the above equality by the commutativity of the first pushout diagram. Hence it induces a map $\widetilde H : Z \times I \longrightarrow Z.$ Though we still have to show that $\widetilde {H}$ is a deformation retraction of $Z$ onto $Z_1$ which is identity on $X_0.$ Do anybody have any idea as to how do I visualize $Z_1$ and $Z_2$ as subspaces of $Z\ $? It is clear that $\widetilde {H}$ is a homotopy relative to $X_0$ which follows from the commutativity of the lower exterior triangle of the later pushout which yields $\widetilde H \big\rvert_{X_0 \times I} = p.$ Now we need to show that $\widetilde H(-,t) \bigg\rvert_{Z_1} = \text {id}_{Z_1},$ for all $t \in I$ and $\widetilde H(z,1) \in Z_1,$ for all $z \in Z.$
RE-EDIT $:$ We can write $$X_0 \cup_F (X × I) = \bigcup\limits_{t \in I} X_0 \cup_{F(a,t) \sim (a,t)} (X \times \{t\}.$$ Then we can say that $$X_0 \cup_{F(a,0) \sim (a,0)} (X \times \{0\}) \subseteq X_0 \cup_F (X \times I).$$ So the homeomorphic copy of $X_0 \cup_{f_1} X$ which is sitting inside $X_0 \cup_F (X \times I)$ is $X_0 \cup_{F(a,0) \sim (a,0) }(X \times \{0\})$ because $$X_0 \cup_{f_1} X \approx X_0 \cup_{f_1 (a) \sim (a,0)} X \approx X_0 \cup_{F(a,0) \sim (a,0)} X.$$ Now since $H(-,t) \big\rvert_{X \times \{0\}} = \text {id}_{X \times \{0\}}$ and $\widetilde H(-,t) \big\rvert_{X_0} = \text {id}_{X_0},$ for all $t \in \Bbb R$ it follows that $\widetilde H \big\rvert_{X_0 \cup_{f_1} X} = \text {id}_{X_0 \cup_{f_1} X}.$ Also since $H(-,0) = \text {id}_{X \times I}$ it follows that $\widetilde H(-,0) = \text {id}_{Z}.$ So the only thing we need to show is that $\widetilde H(z,1) \in X_0 \cup_{f_1} X,$ for all $z \in Z.$
As an adjunction space the identified copy of $X \times I$ sitting inside $Z$ is the image of $X \times I$ and the copy of $X_0$ sitting inside $Z$ is $X_0$ itself because $X_0$ is embedded in $Z.$ We have already seen that at any time $t,$ we have $\widetilde H (x_0,t) = x_0,$ for all $x_0 \in X_0.$ Now in order determine $\widetilde {H} (-,1)$ we need to figure out $H(-,1)$ since the image of $(X \times I) \times I$ in $Z \times I$ followed by $\widetilde H$ factors through $H.$ But we know that the image of $H(-,1)$ lies in $(X \times \{0\}) \cup (A \times I).$ Now if the $H$-image is in $X \times \{0\}$ then it's image in $Z \times I$ will be sent to $X \times \{0\} \cup_{F(a,0) \sim (a,0)} X_0$ by $\widetilde H.$ But the adjunction space $(X \times \{0\}) \cup_{F(a,0) \sim (a,0)} X_0$ is the copy of $X_0 \cup_{f_1} X$ sitting inside $Z.$ So we are done in this case. Now if the $H$-image lies in $A \times I$ then it's image in $Z \times I$ will be sent to $F(a,t) \in X_0$ by $\widetilde H$ (the class of $X_0$ inside $Z$ is identified with itself since $X_0$ is embedded in $Z$) since $F(a,t) \sim (a,t),$ for all $a \in A$ and for all $t \in I.$ But $X_0$ is also embedded inside $X_0 \cup_{f_1} X$ as a pushout and so we are through in this case also. So $\widetilde H$ is a deformation retract of $Z$ onto $Z_1.$ Similarly inverting the role of $F$ by $\overline {F}$ (the reverse homotopy) we get a deformation retract of $Z$ onto $Z_2.$