Is the equivalence between tangent and cotangent bundles canonical for algebraic topologists? Let $M$ be a smooth manifold. Choosing a metric tensor $g$ on $TM$ one gets a vector bundle isomorphism $g^\flat\colon TM\to T^*M$. Changing $g$ we get different maps $g^\flat$.
I would like to understand the statement and its consequences for algebraic topology:

Every two metric choices are homotopic.

My interpretation would be that the metric tensor is a (positive-definite) section $g\colon M\to \mathrm{Sym}^2(T^*M)$ and every two of these are homotopic (through positive-definite sections) by
$$H_t(x) = t g_1(x) + (1-t)g_2(x)$$
Similarily if I treat $g^\flat$ as a section of the bundle $\mathrm{Hom}(TM, T^*M)$ which is an isomorphism on each fiber, the above homotopy yields a homotopy (through isomorphisms as well) between any $g_1^\flat$ and $g_2^\flat$. 
This would mean that from the point of view of algebraic topology, the equivalence between $TM$ and $T^*M$ is "canonical" (induced isomorphisms on (co)homology and similar concepts do not depend on the metric chosen).
Have I got this right?
 A: Being continuously isomorphic is stronger than being homotopy equivalent. The metric, as you say, provides a continuous isomorphism. In this sense, homotopy invariants are 'canonically' the same.
Notice that this is not the same as saying that the map induced on a homotopy invariant by two continuous isomorphisms are the same, but if they differ by a choice of the metric then, as you write, they are indeed the same.
However, this is not the only isomorphism possible between the tangent and the cotangent bundle of the manifold. Suppose your manifold admits a non degenerate closed two form $\omega:M\to \Omega^{2}(M)$, i.e. it is symplectic. Then there is an isomorphism between the tangent and the cotangent bundle induced via $v\mapsto \omega(v,-).$
This is not even a section of the same bundle and as you can check, two choices of the symplectic form are homotopic. Thus they give the same isomorphism on homotopy invariants, but in general different from the one induced by the metric.
Edit: Let me also mention a fact, which escaped me when I originally wrote the answer, which might also be helpful.
In general vector bundles over a manifold (more generally CW complexes) have the homotopy type of the base space. There are many ways to see this, the quickest being that if $p: E\to M$ is a vector bundle with fibre $V$, then the fibration
$$V \to E\xrightarrow{p} M $$
has contractible fiber and so $p_*:\pi_{i}(E)\to \pi_i(M)$ is an isomorphism for all $i>0$ which then implies that $E\simeq M$ as both $E$ and $M$ have the homotopy type of a CW complex.
So the homotopy type of a vector bundle is not that interesting of a question. There are more interesting questions one can ask, such as classifying vector bundles up to isomorphisms which takes us to the realm of topological $K$-theory, which is an interesting topic in and of itself.
