# What does homotopy invariance mean for twisted K-theory?

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?

• 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. May 16 at 8:15
• @WillSawin what do you mean by "homotopy compatible with that isomorphism"? May 16 at 8:25
• 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. May 16 at 8:32