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Tyrone
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Edit Mar. 2020: My original post was more concerned with specific maps than general existence results. Since it perhaps missed the point here is a more worthwhile comment.

No. Let $\alpha$ generate $\pi_6S^3\cong\mathbb{Z}_{12}$ and let $\beta=5\cdot\alpha$. Put $$X=S^3\cup_\alpha S^7,\qquad Y=S^3\cup_\beta S^7.$$$$X=S^3\cup_\alpha e^7,\qquad Y=S^3\cup_\beta e^7.$$ Then $X,Y$ are compact polyhedra and $$X\not\simeq Y.$$ In fact there is no map between these spaces which induces an isomorphism in homology. On the other hand $$X\vee S^3\simeq (S^3\vee S^3)\cup_{(i_1\alpha +i_2\beta)} e^7\simeq Y\vee S^3.$$ Thus $X$ is dominated by $Y\vee S^3$. Clearly $X$ is wedge-indecomposable, but it is neither dominated by $Y$ or by $S^3$.

Replacing $X,Y$ with $\Sigma^nX$ and $\Sigma^n Y$ we obtain examples of suspensions and even stable homotopy types with the same behaviour.

Here is the original answer. No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

Edit Mar. 2020: My original post was more concerned with specific maps than general existence results. Since it perhaps missed the point here is a more worthwhile comment.

No. Let $\alpha$ generate $\pi_6S^3\cong\mathbb{Z}_{12}$ and let $\beta=5\cdot\alpha$. Put $$X=S^3\cup_\alpha S^7,\qquad Y=S^3\cup_\beta S^7.$$ Then $X,Y$ are compact polyhedra and $$X\not\simeq Y.$$ In fact there is no map between these spaces which induces an isomorphism in homology. On the other hand $$X\vee S^3\simeq (S^3\vee S^3)\cup_{(i_1\alpha +i_2\beta)} e^7\simeq Y\vee S^3.$$ Thus $X$ is dominated by $Y\vee S^3$. Clearly $X$ is wedge-indecomposable, but it is neither dominated by $Y$ or by $S^3$.

Replacing $X,Y$ with $\Sigma^nX$ and $\Sigma^n Y$ we obtain examples of suspensions and even stable homotopy types with the same behaviour.

Here is the original answer. No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

Edit Mar. 2020: My original post was more concerned with specific maps than general existence results. Since it perhaps missed the point here is a more worthwhile comment.

No. Let $\alpha$ generate $\pi_6S^3\cong\mathbb{Z}_{12}$ and let $\beta=5\cdot\alpha$. Put $$X=S^3\cup_\alpha e^7,\qquad Y=S^3\cup_\beta e^7.$$ Then $X,Y$ are compact polyhedra and $$X\not\simeq Y.$$ In fact there is no map between these spaces which induces an isomorphism in homology. On the other hand $$X\vee S^3\simeq (S^3\vee S^3)\cup_{(i_1\alpha +i_2\beta)} e^7\simeq Y\vee S^3.$$ Thus $X$ is dominated by $Y\vee S^3$. Clearly $X$ is wedge-indecomposable, but it is neither dominated by $Y$ or by $S^3$.

Replacing $X,Y$ with $\Sigma^nX$ and $\Sigma^n Y$ we obtain examples of suspensions and even stable homotopy types with the same behaviour.

Here is the original answer. No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

Three years later I perhaps understand the question.
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Tyrone
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NoEdit Mar. 2020: My original post was more concerned with specific maps than general existence results. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$Since it perhaps missed the point here is a more worthwhile comment.

No. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree$\alpha$ generate $2$-$\pi_6S^3\cong\mathbb{Z}_{12}$ and degreelet $-1$-self maps included into each respective factor$\beta=5\cdot\alpha$. Now takePut $$X=S^3\cup_\alpha S^7,\qquad Y=S^3\cup_\beta S^7.$$ Then $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold$X,Y$ are compact polyhedra and $$X\not\simeq Y.$$ In fact there is no map between these spaces which induces an isomorphism in homology. ThenOn the other hand $$X\vee S^3\simeq (S^3\vee S^3)\cup_{(i_1\alpha +i_2\beta)} e^7\simeq Y\vee S^3.$$ Thus $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so$X$ is dominated by $S^n$$Y\vee S^3$. Clearly $X$ is homotopywedge-indecomposable, but it is neither dominated by $S^n\vee S^n$$Y$ or by $S^3$. However

Replacing $S^n$$X,Y$ with $\Sigma^nX$ and $\Sigma^n Y$ we obtain examples of suspensions and even stable homotopy types with the same behaviour.

Here is indecomposablethe original answer. No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

Edit Mar. 2020: My original post was more concerned with specific maps than general existence results. Since it perhaps missed the point here is a more worthwhile comment.

No. Let $\alpha$ generate $\pi_6S^3\cong\mathbb{Z}_{12}$ and let $\beta=5\cdot\alpha$. Put $$X=S^3\cup_\alpha S^7,\qquad Y=S^3\cup_\beta S^7.$$ Then $X,Y$ are compact polyhedra and $$X\not\simeq Y.$$ In fact there is no map between these spaces which induces an isomorphism in homology. On the other hand $$X\vee S^3\simeq (S^3\vee S^3)\cup_{(i_1\alpha +i_2\beta)} e^7\simeq Y\vee S^3.$$ Thus $X$ is dominated by $Y\vee S^3$. Clearly $X$ is wedge-indecomposable, but it is neither dominated by $Y$ or by $S^3$.

Replacing $X,Y$ with $\Sigma^nX$ and $\Sigma^n Y$ we obtain examples of suspensions and even stable homotopy types with the same behaviour.

Here is the original answer. No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.

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Tyrone
Tyrone
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No. Take $A=S^n$ and $X=X_1\vee X_2=S^n\vee S^n$. Let $f=2\vee (-1):S^n\rightarrow S^n\vee S^n$ be the sum of the degree $2$- and degree $-1$-self maps included into each respective factor. Now take $g=\nabla:S^n\vee S^n\rightarrow S^n$ to be the fold map. Then $g\circ f\simeq 2-1\simeq 1\simeq id_{S^n}$ so $S^n$ is homotopy dominated by $S^n\vee S^n$. However $S^n$ is indecomposable.