Two definitions of universal hash functions I have seen two definitions of universal hash functions in the literature.

For any $i \geqslant 2$ let $[i]=\{1,\ldots,i\}$.

Definition 1: 


  A family $\mathcal H$
  of hash functions from $[n]$ to $[m]$ is a
  $k$-universal family of hash functions if for any distinct $x_1,\ldots,x_k\in[n]$ and any possibly identical
  $v_1,\ldots,v_k\in[m]$, $$ \Pr\limits_{h\in\mathcal H} [h(x_1)=v_1\wedge \cdots \wedge h(x_k) = v_k] \leqslant 1/m^k \enspace . $$ 


Definition 2:


  A family $\mathcal H$
  of hash functions from $[n]$ to $[m]$ is a
  $k$-universal family of hash functions if for any distinct $x_1,\ldots,x_k\in[n]$, $$ \Pr\limits_{h\in\mathcal H} [h(x_1)=\cdots=h(x_k) ] \leqslant 1/m^{k-1} \enspace . $$ 


In both definitions if equality holds then it is called strongly $k$-universal or alternatively $k$-wise independent hash family.

Clearly the first definition implies the second. But is the other direction true? 

It is easy to see that this is not the case for $n\leqslant m$. 


  This is a counter example when $[n]\subseteq [m]$:
  
  Consider the family $\mathcal H$ composed of only one element, the
  identity function: $x \mapsto x$. It satisfies the second definition
  since for all $x \neq y \in [n], \Pr_{h\in\mathcal H} [h(x) = h(y)] = 0 \leqslant 1/m^{k-1}$. However, it does not satisfy the first definition
  since for all $x,y\in[n], \Pr_{h\in\mathcal H}[h(x) = x \wedge h(y) = y] = 1 &gt; 1/m^k$.
  
  This answer is inspired by this other answer.


However, hash functions are usually employed in applications where $n&gt;m$. So my question is whether this is also the case for $n&gt;m$.
The two definitions are not equivalent.  The second definition does not imply the first.  You can take $\mathcal{H}$ to be the collection of all functions $h$ such that $h(1) = 1$.