A complex vector bundle of rank n can be viewed as a real vector bundle of rank 2n. From nLab, we have that the second Stiefel-Whitney class of the real vector bundle is given by the first Chern class of the complex vector bundle mod 2: $w_2=c_1$ mod 2. Do we have similar relations for other Stiefel-Whitney classes, such as $w_{2n}=c_n$ mod 2?
[Math] Relation between Stiefel-Whitney class and Chern class
algebraic-topologycharacteristic-classes
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The map $\pi_0^*$ fits into a long exact sequence including $$ H^{n-1}(B) \to H^{n-1}(E_0) \to H^n(E, E_0)$$ So, we want to check that $w_{n-1}(\xi_0)$ maps to zero under $H^{n-1}(E_0) \to H^n(E, E_0)$. The class in $H^n(E, E_0)$ is determined by its restrictions to the fibers $H^n(F, F_0)$ which is isomorphic via the corresponding long exact sequence to $H^{n-1}(F_0)$, thus we need to see where $w_{n-1}(\xi_0)$ maps to under the restriction $H^{n-1}(E_0)\to H^{n-1}(F_0)$. At this point, by the definition of $\xi_0$ and naturality, it should be easy to see that $w_{n-1}(\xi_0)$ maps to $w_{n-1}(\tau)$ where $\tau$ is the tangent bundle of the $(n-1)$-sphere.
First of all, your definition of Wu class is incorrect. If $X$ is a closed connected $n$-manifold, there is a unique class $\nu_k \in H^k(X; \mathbb{Z}_2)$ such that for any $x \in H^{n-k}(X; \mathbb{Z}_2)$, $\operatorname{Sq}^k(x) = \nu_k\cup x$. We call $\nu_k$ the $k^{\text{th}}$ Wu class. If $X$ is also smooth, then the Stiefel-Whitney classes of the tangent bundle of $X$ are related to Steenrod squares and Wu classes by the formula
$$w_i = \sum_{k = 0}^i\operatorname{Sq}^k(\nu_{i-k}).$$
Note $\operatorname{Sq}^k(\nu_{i-k})$ is not simply $\nu_k\cup\nu_{i-k}$ unless $i = n$. So we have
\begin{align*} w_1 &= \operatorname{Sq}^0(\nu_1) = \nu_1\\ w_2 &= \operatorname{Sq}^0(\nu_2) + \operatorname{Sq}^1(\nu_1) = \nu_2 + \nu_1\cup\nu_1\\ w_3 &= \operatorname{Sq}^0(\nu_3) + \operatorname{Sq}^1(\nu_2) = \nu_3 + \operatorname{Sq}^1(\nu_2) \end{align*}
It follows that $\nu_1 = w_1$ and $\nu_2 = w_2 + w_1\cup w_1$. However, at this stage we can only deduce $\nu_3 = w_3 + \operatorname{Sq}^1(\nu_2)$. In order to determine $\nu_3$ in terms of Stiefel-Whitney classes, we need to compute $\operatorname{Sq}^1(\nu_2)$. First note that
\begin{align*} \operatorname{Sq}^1(\nu_2) &= \operatorname{Sq}^1(w_2 + w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^1(w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^0(w_1)\cup\operatorname{Sq}^1(w_1) + \operatorname{Sq}^1(w_1)\cup\operatorname{Sq}^0(w_1) && \text{(by Cartan's formula)}\\ &= \operatorname{Sq}^1(w_2) \end{align*}
so $\nu_3 = w_3 + \operatorname{Sq}^1(w_2)$. To compute Steenrod squares of Stiefel-Whitney classes, we use Wu's formula
$$\operatorname{Sq}^i(w_j) = \sum_{t=0}^k\binom{j-i+t-1}{t}w_{i-t}\cup w_{j+t}.$$
In this case, we see that
$$\operatorname{Sq}^1(w_2) = \binom{0}{0}w_1\cup w_2 + \binom{1}{1}w_0\cup w_3 = w_1\cup w_2 + w_3.$$
Therefore, $\nu_3 = w_3 + \operatorname{Sq}^1(w_2) = w_3 + w_1\cup w_2 + w_3 = w_1\cup w_2$. Suppressing the cup symbol, this agrees with the identity given on nLab.
See this note for more details, as well as the computations for $\nu_4$ and $\nu_5$.
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The following is from Lecture notes of Professor Farrell. A hopefully working link is here(page 95): https://www.dropbox.com/s/80n4wd6xctpe6yr/Characteristic%20Classes%20%28Sparkie%20%E7%9A%84%E5%86%B2%E7%AA%81%E5%89%AF%E6%9C%AC%202014-02-19%29.pdf
We now reach proposition 7. Let $E$ be a complex vector bundle, then the mod 2 reduction of the total Chern class of $E$ is the total Stiefel-Whitney class of $E$. Since there is no Chern classes in odd dimensions, we have there is no Stiefel-Whitney classes in odd dimension as well.
$\textbf{Proof}$
Suppose $E$ is a line bundle. Then we know $w_{0}(L)=1, w_{1}(L)=0,w_{2}(L)=e_{\mathbb{Z}_{2}}(L)=\phi(e(L))$. On the other hand we know $c_{0}(L)=1, c_{1}(L)=e(L)$. So this verified it for line bundles.
For sum of line bundles we have $$ E=\oplus^{n}_{i=1}L_{i} $$ So the total Chern class is $$ c(E)=c(L_{1}) \cup \cdots \cup c(L_{n})\mapsto_{\phi}\omega(E)=\omega(L_{1}) \cup \cdots \cup \omega(L_{n}) $$ We are now going to use splitting principle. But there is a subtle point. We now discuss it.
By the splitting principle there exist $$ f:\mathcal{B}\rightarrow B $$ such that $$ f^{*}(E)=\oplus_{i=1}^{n}L_{i} $$ and $$ f^{*}:H^{*}(B,R)\rightarrow H^{*}(\mathcal{B},R) $$ is monic.
Therefore by naturality and the fact $f^{*}$ is monic with respect to $\mathbb{Z}_{2}$ coefficients. $$ \phi(f^{*}(c(E)))=f^{*}(\phi(c(E)))=f^{*}(\omega(E)) $$