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In ordinary K-theory, homotopy invariance means that if $f,g \colon X \to Y$ are homotopic maps then their induced maps on K-theory are equal: $f^* = g^* \colon K(Y) \to K(X)$. My question is how to make sense of this for twisted K-theory. I'll write $K(X,P)$ for the twisted K-theory of $X$ equipped with a 'twist' P; for example, this would be a principal $PU(\mathcal H)$-bundle if we use the definition of Atiyah and Segal.

In twisted K-theory, a map $f \colon X \to Y$ induces a map $K(Y,P) \to K(X,f^*P)$. If $f$ and $g$ are homotopic, we want to be able to say that $f^* = g^*$. The problem is that $f^*$ takes values in $K(X,f^*P)$ and $g^*$ takes values in $K(X,g^*P)$. We know that these two groups are isomorphic because $f^*P \cong g^*P$, but is there are a canonical isomorphism?

So, we can't technically write $f^* = g^*$, but I'm assuming there has to be a canonical isomorphism of twists $f^*P \cong g^*P$ that we can use to identify $K(X,f^*P)$ and $K^*(X, g^*P)$ to make "homotopy invariance" make sense in this context.

Does anyone know where this has been discussed or has an idea of how to construct such a canonical isomorphism?

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  • $\begingroup$ Probably it doesn't have to be a "canonical" isomorphism but rather any isomorphism of twists gives an isomorphism of the two twisted $K$-theories and then $f^* = g^*$ holds if there is a homotopy compatible with that isomorphism. $\endgroup$
    – Will Sawin
    May 16 at 8:15
  • $\begingroup$ @WillSawin what do you mean by "homotopy compatible with that isomorphism"? $\endgroup$
    – Motmot
    May 16 at 8:25
  • $\begingroup$ The homotopy $[0,1]\times X \to Y$ gives a principal bundle on $[0,1] \times X$ whose fibers over $0$ and $1$ are the two bundles. We want to choose a connection on that bundle in the direction along $[0,1]$ so that integrating the connnection from $0$ to $1$ gives the isomorphism. $\endgroup$
    – Will Sawin
    May 16 at 8:32

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