Similar questions seem to have been asked on the stack exchange before, so I apologise for any repetition, but none of the responses seem to answer the question in satisfactory detail, at least for me.
Let $j : Gr^{k,n} \rightarrow Gr^{k,n+1}$ denote the canonical inclusion of Grassmannians, where we take our Grassmannian over our favourite field, say $\mathbb{C}$. Let $\mathcal{T}^{k,n}$ denote the tautological vector subbundle of rank $k$ on $Gr^{k,n}$. To be clear, this means that $\mathcal{T}^{k,n}$ sits inside exact sequence
$0\rightarrow \mathcal{T}^{k,n} \rightarrow \mathcal{O}^{\oplus n}_{Gr^{k,n}} \rightarrow \mathcal{Q}^{k,n} \rightarrow 0$,
which some people call the `universal exact sequence' on the Grassmannian.
I want to prove that $j^{*}(\mathcal{T}^{k,n+1}) \cong \mathcal{T}^{k,n}$.
My thoughts so far:
By definition of pullback, we have
$j^{*}(\mathcal{T}^{k,n+1}) \cong j^{-1}(\mathcal{T}^{k,n+1})\otimes_{j^{-1}\mathcal{O}_{Gr^{k,n+1}}}\mathcal{O}_{Gr^{k,n}}$.
I believe that, in this case, $j^{-1}(O_{Gr^{k,n+1}}) \cong \mathcal{O}_{Gr^{k,n}}$, so that
$j^{*}(\mathcal{T}^{k,n+1}) \cong j^{-1}(\mathcal{T}^{k,n+1})$.
I want to say that $j^{-1}(\mathcal{T}^{k,n+1}) \cong \mathcal{T}^{k,n}$, but I cannot prove this rigorously. I mean, to me at least, this seems very believable, but I just cannot prove it.
Any help would be much appreciated!
Consider the following embedding of vector bundles on $Gr^{k,n}$: $$ \mathcal{T}^{k,n} \to \mathcal{O}_{Gr^{k,n}}^{\oplus n} \to \mathcal{O}_{Gr^{k,n}}^{\oplus (n + 1)}, $$ where the first arrow is the tautological embedding of $Gr^{k,n}$ and the second arrow is an isomorphism onto the first $n$ summands. By the universal property of the Grassmannian it induces a morphism $$ f \colon Gr^{k,n} \to Gr^{k,n+1} $$ such that $f^*\mathcal{T}^{k,n+1} \cong \mathcal{T}^{k,n}$. So, it remains to note that $f$ is precisely the canonical inclusion.