# Tangent bundle and manifold are homotopy equivalent [duplicate]

In a proof I need the following fact:

Let $M$ be a manifold and $TM$ its tangent bundle. Then $M$ and $TM$ are homotopy equivalent and therefore have the same homotopy groups (and explicitly for the sphere $S^n$ we have that $\forall 0 \leq k < n:$ $\pi_k(TS^n) = \pi_k(S^n) = 0$).

I don't necessarily need an exact proof of this claim, more of a sketch. Thank you a lot in advance!

I found the answer to the question I duplicated not enlightening for my purposes so I would like to keep the question...

• Each tangent space is a copy of $\mathbb{R}^n$. Therefore, it is contractible to a point. You can contract each tangent space to the zero point (which is the zero section). – Michael Burr Nov 28 '16 at 15:43
• See the answer to this question. – Michael Albanese Nov 28 '16 at 15:45

Define $f:M \to TM$ to be the the inclusion map (that is, $x \mapsto (x,0)$) and $g:TM \to M$ to be the projection (that is, $(x,v) \mapsto x$).
Clearly, $g \circ f = id_{M}$. Now, we need to show that $f \circ g$ is homotopic to $id_{TM}$. To that end, it suffices to note that each fiber $T_x$ is congruent to $\Bbb R^n$ and therefore contractible.
The fact that it is the tangent bundle is not important. Let $g_0: E\to M$ a vector bundle, $g_0$ is the projection, and let $g_t(z)=g(zt)$ be the homotopy from $g_0$ to $g_1=Id$. Thus the identrity of $E$ is homotopic to the projection (i.e. $g_0\circ i=i$ if $i$ is the inclusion) from $E$ to $M$, and the projection is a homotopy equivalence.