Small question on relative holology if $Y\subset X$ , what is $\ker \delta$ such that $\delta: H_k(X,Y)\rightarrow H_k(Y)$ ?
is it $\ker \delta = H_k(X,Y)$ ?
$\delta$ is the usual connecting homomorphism from the long exact sequence of relative pairs
Please
Thank you.

There's not really much more I can say than is on Wikipedia or can be found in any intro to algebraic topology text. The map $\delta$ which appears in the long exact sequence of a relative pair $(X,Y)$ as $$\cdots \to H_n(Y) \to H_n(X) \stackrel{f_*}{\to} H_n (X,Y) \stackrel{\delta}{\to} H_{n-1}(Y)  \to \cdots$$ is the map which takes a relative cycle $\alpha$ in $H_n(X,Y)$ to its boundary which is in $H_{n-1}(Y)$ because $\alpha$ is a cycle relative to $Y$.
The kernel of $\delta$ is simply those relative cycles which have zero boundary in $Y$. By the exactness of the above sequence, we can also say that the kernal of $\delta$ is equal to the image of $f_*$, which is induced from the usual quotient homomorphism $f\colon C_n(X)\to C_n(X)/C_n(Y)$ appearing in the relative short exact sequence of degree $n$, after passing to homology.
