$dg_y$ carries a subspace of $T_y(Y)$ onto $\mathbb{R}^l$

Question

The question arises from Guillemin and Pallack Page 28 above the frame:

$dg_y$ carries a subspace of $T_y(Y)$ onto $\mathbb{R}^l$ precisely if that subspace and $T_y(Z)$ span all of $T_y(Y)$.

I understand the kernel of $dg_y$ is $T_y(Y)$, but I think $dg_y$ carries a subspace of $T_y(Y)$ onto $\mathbb{R}^l$ precisely if that subspace span all of $T_y(Y)$? Because all $T_y(Z)$ goes to 0 in $\mathbb{R}^l$, which already hit by the kernel of the subspace.....?

Answer 1

This boils down to pure linear algebra.

Claim: If $f:V \to W$ is a surjective linear map with kernel $K$, and $U$ is a subspace of $V$, then $$ f(U) = W \quad \Longleftrightarrow \quad U + K = V. $$

Proof. If $U + K = V$, then $f(U) = f(U + K) = f(V) = W$, as $f$ is surjective. For the converse, since $U + K \leq V$, it suffices to show that the dimensions match. Since $$ \begin{align} \ker f \big|_{U + K} &= \ker f \cap (U + K) \\ &= K \cap (U + K) \\ &= K, \end{align} $$ the rank-nullity theorem gives, $$ \begin{align} \dim V &= \dim W + \dim K \\ &= \dim f(U) + \dim K \\ &= \dim f(U + K) + \dim \ker f \big|_{U + K} \\ &= \dim (U + K). \end{align} $$

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Now, put $V = T_y(Y)$, $W = \Bbb{R}^\ell$, and $f = dg_y$ with $K = T_y(Z)$.