products in the category of banach spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:07:12Z http://mathoverflow.net/feeds/question/65387 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65387/products-in-the-category-of-banach-spaces products in the category of banach spaces Daniel Barter 2011-05-19T01:12:27Z 2011-05-19T02:39:05Z <p>Let <code>$\{X_{\alpha} \}_{\alpha \in A}$</code> be a collection of Banach spaces. It is easy to show that <code>$P = \{(x_{\alpha}) : {\rm sup}_{\alpha} \|x_{\alpha} \| &lt; \infty \}$</code> with <code>$\| (x_{\alpha} ) \| = {\rm sup}_{\alpha} \| x_{\alpha} \|$</code> is a banach space. </p> <p>If the indexing set $A$ is finite, then it is easy to show that $P$ (with the natural projection maps) is the product of <code>$\{X_{\alpha} \}_{\alpha \in A}$</code> in the category of banach spaces.</p> <p>Assume $Y$ is a banach space and $f_{\alpha} : Y \to X_{\alpha}$ is a linear continuous map for each $\alpha$. If <code>${\rm sup}_{\alpha} \|f_{\alpha} \| &lt; \infty$</code> then it is easy to see that the induced linear map $Y \to \Pi X_{\alpha}$ has image a subset of $P$ and is continuous. If this condition is not satisfied then it is not clear to me that there exists a continuous linear map $g:Y \to P$ such that $\pi_{\alpha} \circ g = f_{\alpha}$ for all $\alpha$. </p> <p>Is there any way to prove that $P$ is the categorical product of the collection <code>$\{X_{\alpha} \}_{\alpha \in A}$</code> when $A$ is infinite? </p> http://mathoverflow.net/questions/65387/products-in-the-category-of-banach-spaces/65393#65393 Answer by Yemon Choi for products in the category of banach spaces Yemon Choi 2011-05-19T02:39:05Z 2011-05-19T02:39:05Z <p>[Promoted from comments, as requested]</p> <p>The category of Banach spaces and bounded linear maps does not have arbitrary infinite products. Here is a simple example, using the notation of the question: take each $X_\alpha$ and $Y$ to be a copy of the ground field, fix an unbounded function $\lambda:A\to{\mathbb R}$, and let <code>$f_\alpha: Y \to X_\alpha$</code> be multiplication by <code>$\lambda(\alpha)$</code>. It is then easy to see that the cone <code>$(f_\alpha: Y\to X_\alpha)_{\alpha\in A}$</code> cannot factor through the cone <code>$(\pi_\alpha: P \to X_\alpha)_{\alpha\in A}$</code>. This shows that $P$ is not the product of the $X_\alpha$, and a little more work shows that no such product can exist (assume it does and then show it would have to be isomorphic to $P$).</p>