For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO questionthis related MO question.)
For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes choosingadding a disjoint base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map $$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology).
However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes choosing a base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map $$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology).
However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes adding a disjoint base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map $$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology).
Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempteda modification to make the map look a little more like Bott periodicity.
I think that the answer is a qualified no. It's an interesting map, but as I see On the face of it, itEric's map does not carry the same information as the Bott map. The homotopy information Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in the domainmind, the targetas I understand now, and the map all changeis a result of Snaith that constructs a lotspectrum equivalent to the Bott spectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Of course Snaith's model has been called "Snaith periodicity", with enough new ideas you could try to changebut the map to makeexisting arguments that it is the same are a use and not a proof of Bott map, but I thinkperiodicity. (In that such an effort would besense, Snaith's model is stone soup, although that metaphor is not really fair to his good paper.)
FirstFor context, I think that the Botthere is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx cases For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map $$\Sigma X \to U(2n).$$ Of Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group.
(If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
At first glance, Eric's twisted suspension onlyis very different. It exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. Ifcourse $X$$\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a line bundletotally different homotopy structure from $L$$BU(\infty)$. Moreover, then it has a maptwisted suspensions aren't adjoint to $\mathbb{C}P^\infty$ordinary delooping. Eric thus constructs a map $$\Sigma^L X \to \mathbb{C}P^\infty,$$ which means that Instead, the twisted suspension also hasspace of maps $\Sigma^L X \to Y$ is adjoint to sections of a line bundle over $L$. You can check that$X$ with fiber $\Sigma^L X$$\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. Moreover
However, the purpose of the suspension maps in spectraSnaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott periodicityelement $\beta$. (I think that the "+" subscript just denotes choosing a base point.) You can see what is to gocoming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a mapCW complex $\Sigma X \to Y$ to its adjoint map$K$ are just the rational homology $X \to \mathcal{L}Y$$H_*(K,\mathbb{Q})$. In this case, since it (This is arelated to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension, you instead get a bundle over become the same. So Snaith's model is built from the fact that the homology of $X$ with fiber$\mathbb{C}P^\infty$ equals the homotopy of $\mathcal{L}^2\mathbb{C}P^\infty$$BU(\infty)$. Again, this could be interesting Moreover, but itthere is not Bott periodicity.
Finally Eric's step 3 formally desuspends both sides to getan important determinant map $$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty \to \Sigma^{-2}\mathbb{C}P^\infty.$$$$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ You can't desuspend unless you first suspend infinitely many timesthat takes the direct sum operation for bundles to get a spectrumtensor multiplication of line bundles. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on Snaith makes a yet different trackmoral inverse to this map (and not just in rational homology). Without changes
Still, thissearching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a really complicated spectrum reminiscentpurely algebraic proof of the sphere spectrumfundamental theorem of algebra. If $\mathbb{C}P^\infty$ is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ is too complicated to resemble it. Edit: Eric points out that Snaith actually established The fundamental theorem of algebra is not a way to modify $\Sigma_+^{\infty}\mathbb{C}P^\infty$ (the plus sign has to dopurely algebraic statement! It is an analytic theorem with choosing base pointsan algebraic conclusion, according to one reference) to make itsince the complex numbers are defined analytically. The best you can do is a mostly algebraic proof, using some minimal analytic information such as that $BU(\infty)$$\mathbb{R}$ is real-closed using the intermediate value theorem. What Likewise, Bott periodicity is not clear to mea purely homotopy-theoretic theorem; it is whether this resulta Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof of Bott periodicity or athat carefully uses as little Lie theory as possible. The proof by Bruno Harris fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use of Bott periodicityabout the unitary groups.
(The answer is significantly revised now that I know more about Snaith's result.)
Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempted modification to make the map look a little more like Bott periodicity.
I think that the answer is no. It's an interesting map, but as I see it, it does not carry the same information as the Bott map. The homotopy information in the domain, the target, and the map all change a lot. Of course, with enough new ideas you could try to change the map to make it the Bott map, but I think that such an effort would be stone soup.
First, I think that the Bott map as Bott constructed it in the Annals is beautiful. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx cases, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map $$\Sigma X \to U(2n).$$ Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group.
(If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
Eric's twisted suspension only exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. If $X$ is a space with a line bundle $L$, then it has a map to $\mathbb{C}P^\infty$. Eric thus constructs a map $$\Sigma^L X \to \mathbb{C}P^\infty,$$ which means that the twisted suspension also has a line bundle $L$. You can check that $\Sigma^L X$ depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. Moreover, the purpose of the suspension maps in spectra and Bott periodicity is to go from a map $\Sigma X \to Y$ to its adjoint map $X \to \mathcal{L}Y$. In this case, since it is a twisted suspension, you instead get a bundle over $X$ with fiber $\mathcal{L}^2\mathbb{C}P^\infty$. Again, this could be interesting, but it is not Bott periodicity.
Finally Eric's step 3 formally desuspends both sides to get
$$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty \to \Sigma^{-2}\mathbb{C}P^\infty.$$
You can't desuspend unless you first suspend infinitely many times to get a spectrum. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on a yet different track. Without changes, this is a really complicated spectrum reminiscent of the sphere spectrum. If $\mathbb{C}P^\infty$ is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ is too complicated to resemble it. Edit: Eric points out that Snaith actually established a way to modify $\Sigma_+^{\infty}\mathbb{C}P^\infty$ (the plus sign has to do with choosing base points, according to one reference) to make it $BU(\infty)$. What is not clear to me is whether this result is a proof of Bott periodicity or a use of Bott periodicity.
Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is a modification to make the map look more like Bott periodicity.
I think that the answer is a qualified no. On the face of it, Eric's map does not carry the same information as the Bott map. Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in mind, as I understand now, is a result of Snaith that constructs a spectrum equivalent to the Bott spectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Snaith's model has been called "Snaith periodicity", but the existing arguments that it is the same are a use and not a proof of Bott periodicity. (In that sense, Snaith's model is stone soup, although that metaphor is not really fair to his good paper.)
For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
At first glance, Eric's twisted suspension is very different. It exists for $\mathbb{C}P^\infty = BU(1)$, and of course $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a totally different homotopy structure from $BU(\infty)$. Moreover, twisted suspensions aren't adjoint to ordinary delooping. Instead, the space of maps $\Sigma^L X \to Y$ is adjoint to sections of a bundle over $X$ with fiber $\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed.
However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes choosing a base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map $$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology).
Still, searching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a purely algebraic proof of the fundamental theorem of algebra. The fundamental theorem of algebra is not a purely algebraic statement! It is an analytic theorem with an algebraic conclusion, since the complex numbers are defined analytically. The best you can do is a mostly algebraic proof, using some minimal analytic information such as that $\mathbb{R}$ is real-closed using the intermediate value theorem. Likewise, Bott periodicity is not a purely homotopy-theoretic theorem; it is a Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof that carefully uses as little Lie theory as possible. The proof by Bruno Harris fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use about the unitary groups.
(The answer is significantly revised now that I know more about Snaith's result.)