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Maxime Ramzi
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In homotopy theory, this happens when you use stable splittings of spaces to analyze homotopy types. For example, (writing $X_+$ for $X$ with a disjoint basepoint), $X_+ \not\simeq X \vee S^0$ (as pointed spaces, generally), but $$ \Sigma ( X_+ ) \simeq \Sigma X \vee S^1 \simeq \Sigma ( X_+ \vee S^0 ). $$ Also $\Sigma (X\wedge Y) \simeq (\Sigma X) \vee Y \simeq X \wedge \Sigma Y$$\Sigma (X\wedge Y) \simeq (\Sigma X) \wedge Y \simeq X \wedge \Sigma Y$ (since $\Sigma X = S^1 \wedge X$ and $\wedge$ is commutative and associative). From this we get, for example $$ \Sigma ( (X\times Y)_+ ) = \Sigma( X_+ \wedge Y_+) \simeq \Sigma ( (X\vee S^0) \wedge (Y\vee S^0) ) = \Sigma ((( X\vee Y \vee (X \wedge Y))_+). $$ This is a pretty painless way to show the stable splitting of products. Another simple formula that is useful for this kind of argument is the James splitting $$ \Sigma ( \Omega \Sigma X) \simeq \Sigma \textstyle\left( \bigvee_{n\geq 0} X^{\wedge n} \right). $$ This is used by B. Gray, for example, in his nice proof of the Hilton-Milnor theorem.

In homotopy theory, this happens when you use stable splittings of spaces to analyze homotopy types. For example, (writing $X_+$ for $X$ with a disjoint basepoint), $X_+ \not\simeq X \vee S^0$ (as pointed spaces, generally), but $$ \Sigma ( X_+ ) \simeq \Sigma X \vee S^1 \simeq \Sigma ( X_+ \vee S^0 ). $$ Also $\Sigma (X\wedge Y) \simeq (\Sigma X) \vee Y \simeq X \wedge \Sigma Y$ (since $\Sigma X = S^1 \wedge X$ and $\wedge$ is commutative and associative). From this we get, for example $$ \Sigma ( (X\times Y)_+ ) = \Sigma( X_+ \wedge Y_+) \simeq \Sigma ( (X\vee S^0) \wedge (Y\vee S^0) ) = \Sigma ((( X\vee Y \vee (X \wedge Y))_+). $$ This is a pretty painless way to show the stable splitting of products. Another simple formula that is useful for this kind of argument is the James splitting $$ \Sigma ( \Omega \Sigma X) \simeq \Sigma \textstyle\left( \bigvee_{n\geq 0} X^{\wedge n} \right). $$ This is used by B. Gray, for example, in his nice proof of the Hilton-Milnor theorem.

In homotopy theory, this happens when you use stable splittings of spaces to analyze homotopy types. For example, (writing $X_+$ for $X$ with a disjoint basepoint), $X_+ \not\simeq X \vee S^0$ (as pointed spaces, generally), but $$ \Sigma ( X_+ ) \simeq \Sigma X \vee S^1 \simeq \Sigma ( X_+ \vee S^0 ). $$ Also $\Sigma (X\wedge Y) \simeq (\Sigma X) \wedge Y \simeq X \wedge \Sigma Y$ (since $\Sigma X = S^1 \wedge X$ and $\wedge$ is commutative and associative). From this we get, for example $$ \Sigma ( (X\times Y)_+ ) = \Sigma( X_+ \wedge Y_+) \simeq \Sigma ( (X\vee S^0) \wedge (Y\vee S^0) ) = \Sigma ((( X\vee Y \vee (X \wedge Y))_+). $$ This is a pretty painless way to show the stable splitting of products. Another simple formula that is useful for this kind of argument is the James splitting $$ \Sigma ( \Omega \Sigma X) \simeq \Sigma \textstyle\left( \bigvee_{n\geq 0} X^{\wedge n} \right). $$ This is used by B. Gray, for example, in his nice proof of the Hilton-Milnor theorem.

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Jeff Strom
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In homotopy theory, this happens when you use stable splittings of spaces to analyze homotopy types. For example, (writing $X_+$ for $X$ with a disjoint basepoint), $X_+ \not\simeq X \vee S^0$ (as pointed spaces, generally), but $$ \Sigma ( X_+ ) \simeq \Sigma X \vee S^1 \simeq \Sigma ( X_+ \vee S^0 ). $$ Also $\Sigma (X\wedge Y) \simeq (\Sigma X) \vee Y \simeq X \wedge \Sigma Y$ (since $\Sigma X = S^1 \wedge X$ and $\wedge$ is commutative and associative). From this we get, for example $$ \Sigma ( (X\times Y)_+ ) = \Sigma( X_+ \wedge Y_+) \simeq \Sigma ( (X\vee S^0) \wedge (Y\vee S^0) ) = \Sigma ((( X\vee Y \vee (X \wedge Y))_+). $$ This is a pretty painless way to show the stable splitting of products. Another simple formula that is useful for this kind of argument is the James splitting $$ \Sigma ( \Omega \Sigma X) \simeq \Sigma \textstyle\left( \bigvee_{n\geq 0} X^{\wedge n} \right). $$ This is used by B. Gray, for example, in his nice proof of the Hilton-Milnor theorem.

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